Mechanics Alıştırma Soruları — JAMB UTME

Soru 1Soru →

A U-tube open at both ends contains mercury of density $13\text{}600\text{ kg/m}^3$ . Water of density $1000\text{ kg/m}^3$ is poured into one arm until the water column reaches a height of $27.2\text{ cm}$ . What is the difference in height, in $\text{cm}$ , between the mercury surfaces in the two arms?

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Cevap: 2

Cevap

The difference in height between the mercury surfaces in the two arms is

2.0\text{ cm}

.

At the boundary level where water meets mercury, the pressure produced by the

27.2\text{ cm}

water column must equal the pressure of the mercury column above that same horizontal level. Using

h_w \rho_w = h_m \rho_m

, we solve for the mercury height difference:

h_m = \frac{27.2 \times 1000}{13600} = 2.0\text{ cm}

.

Adım Adım Çözüm

1

Equate the hydrostatic pressure exerted by the water column to the hydrostatic pressure exerted by the balancing mercury column at the interface level.

h_w \rho_w g = h_m \rho_m g

At the same horizontal level within a continuous fluid at rest, the pressures must be equal.

2

Cancel the acceleration due to gravity (

g

) from both sides of the equation.

h_w \rho_w = h_m \rho_m

Gravity acts equally on both liquid columns.

3

Substitute the known values (

h_w = 27.2\text{ cm}

,

\rho_w = 1000\text{ kg/m}^3

,

\rho_m = 13600\text{ kg/m}^3

) into the pressure relation.

27.2 \times 1000 = h_m \times 13600

Inserting the physical quantities isolates the unknown mercury column height

h_m

.

4

Solve for the height difference

h_m

of the mercury levels.

h_m = \frac{27200}{13600} = 2.0\text{ cm}

Dividing the water pressure head product by the density of mercury yields the height of the mercury column.

Anahtar Kavram

Hydrostatic pressure equilibrium in immiscible fluids (U-tube manometer)

Soru 2Soru →

A solid block of mass $0.5\text{ kg}$ is completely immersed in water and displaces $0.2\text{ kg}$ of water. What is the magnitude of the upthrust exerted by the water on the block? (Take $g = 10\text{ m/s}^2$ )

2\text{ N}

3\text{ N}

5\text{ N}

7\text{ N}

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Cevap: $2\text{ N}$

Cevap

The magnitude of the upthrust exerted on the block is

2\text{ N}

.

By Archimedes' principle, the buoyant force (upthrust) acting on a submerged object equals the weight of the liquid displaced by the object. Since the mass of displaced water is

0.2\text{ kg}

and

g = 10\text{ m/s}^2

, the weight of the displaced water is

0.2\text{ kg} \times 10\text{ m/s}^2 = 2\text{ N}

.

Adım Adım Çözüm

1

Identify Archimedes' Principle

Upthrust (

U

) is equal to the weight of the fluid displaced by the immersed body.

Archimedes' principle states that the buoyant force on a submerged body equals the weight of the fluid it displaces.

2

Calculate the weight of the displaced water

W_{\text{displaced}} = m_{\text{water}} \times g = 0.2\text{ kg} \times 10\text{ m/s}^2 = 2\text{ N}

Weight is calculated as mass multiplied by acceleration due to gravity.

3

State the upthrust

U = 2\text{ N}

The upthrust is directly equal to the weight of the displaced water.

Anahtar Kavram

Archimedes' Principle and Upthrust

Tahmini Süre:45s

Soru 3Soru →

The aerodynamic drag force $F$ acting on an object moving through a fluid of density $\rho$ with cross-sectional area $A$ at speed $v$ is modeled by the equation $F = \frac{1}{2} C_d \rho^a A^b v^c$ , where $C_d$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the exponent $c$ ?

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Cevap: 2

Cevap

The numerical value of the exponent

c

is 2.

By applying the principle of dimensional homogeneity, the base dimension of time on the left side is

\text{T}^{-2}

(from force

[F] = \text{M L T}^{-2}

). On the right side, the only quantity containing time is velocity

[v] = \text{L T}^{-1}

, raised to power

c

, giving

\text{T}^{-c}

. Equating the exponents gives

-2 = -c

, so

c = 2

.

Adım Adım Çözüm

1

Identify the base dimensions of each physical quantity in the given equation.

[F] = \text{M L T}^{-2}

,

[\rho] = \text{M L}^{-3}

,

[A] = \text{L}^2

, and

[v] = \text{L T}^{-1}

.

C_d

is dimensionless (

[C_d] = 1

).

Dimensional homogeneity requires both sides of a physical equation to have identical base dimensions.

2

Substitute the base dimensions into the formula

F = \frac{1}{2} C_d \rho^a A^b v^c

and simplify.

\text{M L T}^{-2} = (\text{M L}^{-3})^a (\text{L}^2)^b (\text{L T}^{-1})^c = \text{M}^a \text{L}^{-3a + 2b + c} \text{T}^{-c}.

Combining powers of base dimensions allows direct comparison of corresponding exponents.

3

Equate the exponent of time (T) on both sides of the dimensional equation.

-2 = -c \implies c = 2.

The exponent of T on the left side is -2, which must equal the exponent of T on the right side (-c).

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:1m 15s

Soru 4Soru →

A particle starts from rest and accelerates uniformly at a rate of $4\text{ m/s}^2$ for a duration $t_1$ . Immediately after reaching its maximum velocity, it decelerates uniformly at $2\text{ m/s}^2$ until coming to rest. If the total distance covered during the entire motion is $600\text{ m}$ , what is the total time of motion in seconds?

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Cevap: 30

Cevap

The total time of motion is 30 seconds.

For a two-stage motion starting and ending at rest, the peak velocity is

v_{\text{max}} = a_1 t_1 = a_2 t_2

, giving a time ratio

t_2 / t_1 = a_1 / a_2 = 4 / 2 = 2

. The total distance

S

is the area under the velocity-time triangle,

S = \frac{1}{2} v_{\text{max}} (t_1 + t_2) = \frac{1}{2} (4 t_1) (3 t_1) = 6 t_1^2

. Setting

6 t_1^2 = 600

yields

t_1 = 10\text{ s}

, which gives a total time

T = t_1 + t_2 = 30\text{ s}

.

Adım Adım Çözüm

1

Relate maximum velocity to the acceleration time

t_1

v_{\text{max}} = 4 t_1

Using

v = u + a t

starting from rest (

u = 0

).

2

Relate deceleration time

t_2

to

t_1

t_2 = 2 t_1

The final velocity is

0

, so

0 = v_{\text{max}} - a_2 t_2 \implies t_2 = \frac{4 t_1}{2} = 2 t_1

.

3

Express the total displacement

S

as a function of

t_1

S = 6 t_1^2

Displacement during acceleration

s_1 = \frac{1}{2}(4)t_1^2 = 2 t_1^2

. Displacement during deceleration

s_2 = \frac{1}{2}(2)(2 t_1)^2 = 4 t_1^2

. Total

S = 2 t_1^2 + 4 t_1^2 = 6 t_1^2

.

4

Solve for the acceleration time

t_1

t_1 = 10\text{ s}

Given

S = 600\text{ m}

, we have

6 t_1^2 = 600 \implies t_1^2 = 100 \implies t_1 = 10\text{ s}

.

5

Calculate the total time of motion

T

T = 30\text{ s}

Total time is the sum of both phases:

T = t_1 + t_2 = t_1 + 2 t_1 = 3 t_1 = 3(10) = 30\text{ s}

.

Anahtar Kavram

Multi-stage uniform motion and average velocity relations

Tahmini Süre:3m 0s

Soru 5Soru →

A body of mass $0.4\text{ kg}$ suspended vertically from a helical spring produces a static extension of $0.1\text{ m}$ . The body is then pulled down further and set into vertical simple harmonic motion with an amplitude of $0.05\text{ m}$ . What is the maximum velocity of the body in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

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Cevap: 0.5

Cevap

The maximum velocity of the body during oscillation is

0.5\text{ m/s}

.

At static equilibrium, weight balances restoring force (

mg = ke

), giving

\frac{k}{m} = \frac{g}{e} = \frac{10}{0.1} = 100\text{ s}^{-2}

. The angular frequency is

\omega = \sqrt{\frac{k}{m}} = 10\text{ rad/s}

. In SHM, the maximum velocity occurs at the central equilibrium position and is given by

v_{\max} = \omega A = 10 \times 0.05 = 0.5\text{ m/s}

.

Adım Adım Çözüm

1

Relate spring stiffness to static extension

\frac{k}{m} = 100\text{ s}^{-2}

At vertical static equilibrium, the weight of the mass equals the restoring force:

mg = ke \implies \frac{k}{m} = \frac{g}{e} = \frac{10\text{ m/s}^2}{0.1\text{ m}} = 100\text{ s}^{-2}

.

2

Determine the angular frequency

\omega = 10\text{ rad/s}

The angular frequency of a mass-spring system is given by

\omega = \sqrt{\frac{k}{m}} = \sqrt{100} = 10\text{ rad/s}

.

3

Calculate maximum velocity

v_{\max} = 0.5\text{ m/s}

The maximum speed in simple harmonic motion occurs at the equilibrium position and is computed using

v_{\max} = \omega A = 10\text{ rad/s} \times 0.05\text{ m} = 0.5\text{ m/s}

.

Anahtar Kavram

Maximum velocity and angular frequency derived from static extension in Simple Harmonic Motion

Soru 6Soru →

A car traveling along a straight road at an initial speed of $20\text{ m/s}$ applies its brakes, causing a uniform deceleration of $5\text{ m/s}^2$ until it comes to a complete stop. What is the total distance traveled by the car during this braking period?

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Cevap: 40

Cevap

The total distance traveled by the car while coming to a stop is

40\text{ m}

.

Using the third equation of linear motion

v^2 = u^2 + 2as

with

u = 20\text{ m/s}

,

v = 0\text{ m/s}

, and acceleration

a = -5\text{ m/s}^2

, we obtain

0 = 400 - 10s

, which simplifies directly to

s = 40\text{ m}

.

Adım Adım Çözüm

1

Identify the given kinematic parameters from the problem statement.

u = 20\text{ m/s}

,

v = 0\text{ m/s}

,

a = -5\text{ m/s}^2

The car decelerates to a stop, so final velocity is zero and acceleration is negative relative to initial direction of motion.

2

Select the appropriate equation of motion linking

u

,

v

,

a

, and displacement

s

.

v^2 = u^2 + 2as

This equation directly relates initial velocity, final velocity, acceleration, and distance without requiring time.

3

Substitute the known values into the equation and solve for distance

s

.

0^2 = 20^2 + 2(-5)s \implies 10s = 400 \implies s = 40\text{ m}

Algebraic simplification yields the stopping distance.

Anahtar Kavram

Uniformly Accelerated Motion and Stopping Distance

Soru 7Soru →

A car starts from rest and accelerates uniformly to a velocity of $16\text{ m/s}$ in $4\text{ s}$ . It continues at this constant velocity for $8\text{ s}$ , and then decelerates uniformly to rest in another $4\text{ s}$ . What is the average speed of the car for the entire journey?

12\text{ m/s}

16\text{ m/s}

8\text{ m/s}

24\text{ m/s}

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Cevap: $12\text{ m/s}$

Cevap

12\text{ m/s}

The average speed of an object undergoing multi-stage motion is defined as the total distance traveled divided by the total time taken. The total distance covered is

32\text{ m}

(during acceleration)

+ 128\text{ m}

(during constant velocity)

+ 32\text{ m}

(during deceleration)

= 192\text{ m}

. Dividing

192\text{ m}

by the total time of

16\text{ s}

yields

12\text{ m/s}

.

Adım Adım Çözüm

1

Calculate the distance covered in each stage of motion.

Acceleration stage (

s_1

):

\frac{0 + 16}{2} \times 4 = 32\text{ m}

. Constant velocity stage (

s_2

):

16 \times 8 = 128\text{ m}

. Deceleration stage (

s_3

):

\frac{16 + 0}{2} \times 4 = 32\text{ m}

.

The total distance is the sum of distances traveled during acceleration, constant speed, and deceleration.

2

Find total distance traveled and total time taken.

Total distance (

S

) =

32\text{ m} + 128\text{ m} + 32\text{ m} = 192\text{ m}

. Total time (

T

) =

4\text{ s} + 8\text{ s} + 4\text{ s} = 16\text{ s}

.

Average speed requires total distance divided by total elapsed time.

3

Compute the average speed.

Average speed =

\frac{S}{T} = \frac{192\text{ m}}{16\text{ s}} = 12\text{ m/s}

.

Dividing total distance by total time gives the average speed over the entire motion.

Anahtar Kavram

Average Speed in Multi-Stage Motion

Tahmini Süre:1m 30s

Soru 8Soru →

A ball $P$ is dropped from rest from the top of a cliff of height $100\text{ m}$ . At the same instant, another ball $Q$ is projected vertically upwards from the base of the cliff along the same vertical line with an initial speed of $50\text{ m/s}$ . Taking $g = 10\text{ m/s}^2$ , at what height above the ground do the two balls meet?

20\text{ m}

50\text{ m}

80\text{ m}

100\text{ m}

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Cevap: $80\text{ m}$

Cevap

The two balls meet at a height of

80\text{ m}

above the ground.

Equating the position equations of both objects gives

100 - 5t^2 = 50t - 5t^2

, which simplifies to

50t = 100

, so

t = 2\text{ s}

. Substituting

t = 2\text{ s}

into the vertical height formula gives

h = 80\text{ m}

above the ground.

Adım Adım Çözüm

1

Set up equations of motion for both balls in terms of time

t

.

For ball

P

(dropped from top): height above ground

h_P = 100 - \frac{1}{2}gt^2 = 100 - 5t^2

.
For ball

Q

(projected from ground): height above ground

h_Q = ut - \frac{1}{2}gt^2 = 50t - 5t^2

.

Both balls move simultaneously under gravity along the same vertical line.

2

Equate the heights

h_P = h_Q

to find the time of meeting

t

.

100 - 5t^2 = 50t - 5t^2 \implies 50t = 100 \implies t = 2\text{ s}

.

The balls pass each other when their heights above the ground are equal.

3

Calculate the height above ground using

t = 2\text{ s}

.

h = 50(2) - 5(2)^2 = 100 - 20 = 80\text{ m}

.

Substituting

t = 2\text{ s}

into either height expression yields the position where they meet.

Anahtar Kavram

Relative vertical motion under gravity

Alternatif Yöntem

Using relative velocity: The relative acceleration between the two balls is

g - g = 0\text{ m/s}^2

. The relative speed of approach is constant at

50\text{ m/s}

. The initial separation is

100\text{ m}

, so time to meet is

t = \frac{100}{50} = 2\text{ s}

. Height above ground is then

h = 100 - \frac{1}{2}(10)(2)^2 = 80\text{ m}

.

Tahmini Süre:1m 30s

Soru 9Soru →

A body is projected horizontally from the top of a cliff $45\text{ m}$ high. If it lands on flat ground at a horizontal distance of $120\text{ m}$ from the base of the cliff, what is the speed of the body just before it strikes the ground? (Take $g = 10\text{ m/s}^2$ )

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Cevap: 50

Cevap

The speed of the body just before striking the ground is 50 m/s.

The time of fall is determined by the height of

45\text{ m}

, yielding

t = \sqrt{2h/g} = 3\text{ s}

. The horizontal speed is constant at

120 / 3 = 40\text{ m/s}

. The vertical velocity gained on impact is

v_y = gt = 30\text{ m/s}

. Combining these mutually perpendicular velocity components gives a final impact speed of

v = \sqrt{40^2 + 30^2} = 50\text{ m/s}

.

Adım Adım Çözüm

1

Calculate the time of flight from the vertical height

t = 3 s

Vertical acceleration is constant under gravity while initial vertical velocity is zero.

2

Compute the constant horizontal component of velocity

v_x = 40 m/s

Horizontal speed is uniform because zero horizontal force acts on the projectile.

3

Compute the final vertical component of velocity at impact

v_y = 30 m/s

Vertical speed increases linearly with time due to gravitational acceleration.

4

Determine the magnitude of the resultant velocity vector

v = 50 m/s

The horizontal and vertical components are perpendicular, so their vector sum uses the Pythagorean theorem.

Anahtar Kavram

Horizontal Projection and Impact Velocity Vector

Tahmini Süre:1m 30s

Soru 10Soru →

An athlete throws a javelin from ground level such that its initial vertical component of velocity is $40\text{ m/s}$ and its initial horizontal component of velocity is $30\text{ m/s}$ . What is the horizontal distance in metres covered by the javelin when it reaches a height of $35\text{ m}$ above the ground for the first time? (Take $g = 10\text{ m/s}^2$ )

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Cevap: 30

Cevap

The horizontal distance covered by the javelin when it reaches a height of 35 m for the first time is 30 m.

Using

y = u_y t - \frac{1}{2}g t^2

with

u_y = 40\text{ m/s}

,

y = 35\text{ m}

, and

g = 10\text{ m/s}^2

yields

35 = 40t - 5t^2

. Dividing by 5 gives

t^2 - 8t + 7 = 0

, which factors to

(t - 1)(t - 7) = 0

. The roots are

t = 1\text{ s}

(ascent) and

t = 7\text{ s}

(descent). For the first time,

t = 1\text{ s}

. The horizontal displacement is

x = u_x t = 30\text{ m/s} \times 1\text{ s} = 30\text{ m}

.

Adım Adım Çözüm

1

Set up the vertical motion equation to find the time when height is 35 m

35 = 40t - 5t^2

Vertical displacement in projectile motion depends on the vertical initial velocity component and acceleration due to gravity.

2

Solve the quadratic equation for time

t

t^2 - 8t + 7 = 0 \implies t = 1\text{ s} \text{ or } t = 7\text{ s}

A projectile reaches a given non-peak height twice: once ascending and once descending.

3

Select the first time value and calculate horizontal distance

x = u_x \times t = 30 \times 1 = 30\text{ m}

Horizontal velocity remains constant throughout the flight, so distance is speed multiplied by time.

Anahtar Kavram

Independence of vertical and horizontal components in projectile motion

Soru 11Soru →

An object is projected from ground level at an angle of $60^\circ$ to the horizontal. If the horizontal component of its initial velocity is $25\text{ m/s}$ , calculate the maximum height reached by the object in meters. (Take $g = 10\text{ m/s}^2$ )

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Cevap: 93.75

Cevap

The maximum height reached by the object is

93.75\text{ m}

.

The maximum vertical height attained by a projectile depends on its vertical velocity component

u_y = u \sin \theta

. Resolving the initial velocity gives

u = 50\text{ m/s}

and

u_y = 25\sqrt{3}\text{ m/s}

. Substituting into

H = \frac{u_y^2}{2g}

yields

93.75\text{ m}

.

Adım Adım Çözüm

1

Find the magnitude of the initial velocity

u = 50\text{ m/s}

The horizontal velocity component remains constant throughout flight and is given by

u_x = u \cos \theta

.

2

Calculate the initial vertical velocity component

u_y = 25\sqrt{3}\text{ m/s}

Vertical component of velocity is calculated using

u_y = u \sin \theta

.

3

Calculate the maximum height

H = 93.75\text{ m}

At maximum height, vertical velocity is zero, giving

H = \frac{u_y^2}{2g}

.

Anahtar Kavram

Resolution of velocity components in projectile motion and calculation of maximum height

Soru 12Soru →

When a stationary object in an isolated system explodes into two fragments of unequal mass, the fragment with the larger mass acquires a greater magnitude of linear momentum than the fragment with the smaller mass.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is false. Both fragments acquire linear momentum of equal magnitude in opposite directions.

The statement is false because conservation of linear momentum requires the total initial momentum (zero) to equal the total final momentum. Therefore, the two fragments move in opposite directions with linear momenta of equal magnitude, regardless of their masses.

Adım Adım Çözüm

1

Analyze internal forces acting during the explosion

By Newton's third law, the force exerted on the first fragment is equal in magnitude and opposite in direction to the force exerted on the second fragment (

F_1 = -F_2

).

Explosive forces are internal action-reaction pairs.

2

Apply the impulse-momentum theorem

Since both forces act for the exact same duration

\Delta t

, the impulse

J_1 = F_1 \Delta t

equals

-J_2 = -F_2 \Delta t

. Therefore, the change in linear momentum

\Delta p_1 = -\Delta p_2

.

Impulse delivered to an object equals its change in linear momentum.

3

Compare momentum magnitudes

|p_1| = |p_2|

, which means

m_1 v_1 = m_2 v_2

.

Initial momentum was zero (

p_{\text{initial}} = 0

), so the sum of final momentum vectors must be zero (

p_1 + p_2 = 0

).

Anahtar Kavram

Conservation of Linear Momentum and Newton's Third Law

Soru 13Soru →

A mountain climber of mass $60\text{ kg}$ climbs a vertical height of $15\text{ m}$ in a time of $30\text{ s}$ . Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , calculate the average power expended by the climber in Watts.

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Cevap: 300

Cevap

The average power expended by the climber is

300\text{ W}

.

The work done in lifting a mass

m

through vertical height

h

is given by

W = mgh = 60 \times 10 \times 15 = 9000\text{ J}

. The average power is the rate of doing work,

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Adım Adım Çözüm

1

Identify the given values and formula for work done against gravity.

Mass

m = 60\text{ kg}

, vertical displacement

h = 15\text{ m}

, acceleration due to gravity

g = 10\text{ m s}^{-2}

, and time

t = 30\text{ s}

. Work done formula:

W = mgh

.

The climber works against gravity to increase their potential energy by an amount equal to

mgh

.

2

Calculate the total work done.

W = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}

.

Multiplying force (

mg

) by vertical distance (

h

) yields work done in Joules.

3

Calculate the average power.

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Power is defined as work done divided by the time interval (

P = \frac{W}{t}

).

Anahtar Kavram

Power as the rate of doing work against gravity

Tahmini Süre:45s

Soru 14Soru →

A body is projected vertically upwards from the ground with an initial velocity $u$ . It passes a point at a height of $40\text{ m}$ above the ground at $t = 2\text{ s}$ while ascending and again at $t = 4\text{ s}$ while descending. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the initial speed of projection $u$ of the body?

20\text{ m/s}

30\text{ m/s}

40\text{ m/s}

60\text{ m/s}

Cevabı ve açıklamayı göster

Cevap: $30\text{ m/s}$

Cevap

The initial speed of projection of the body is

30\text{ m/s}

.

The position of a body thrown vertically upwards is given by

h = ut - \frac{1}{2}gt^2

. Rearranging this equation into standard quadratic form gives

t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

. The roots

t_1

and

t_2

correspond to the times the body reaches height

h

. By Vieta's formulas, the sum of the times is

t_1 + t_2 = \frac{2u}{g}

. Substituting

t_1 = 2\text{ s}

,

t_2 = 4\text{ s}

, and

g = 10\text{ m/s}^2

gives

6 = \frac{2u}{10}

, which solves to

u = 30\text{ m/s}

. Alternatively, the time to reach maximum height is the midpoint

t_{\text{top}} = \frac{t_1 + t_2}{2} = 3\text{ s}

, and at maximum height

v = 0 = u - gt_{\text{top}}

, yielding

u = 10 \times 3 = 30\text{ m/s}

.

Adım Adım Çözüm

1

Set up the vertical motion displacement equation

h = ut - \frac{1}{2}gt^2

The equation describes the vertical position

h

at any time

t

for a projectile launched from ground level with initial speed

u

.

2

Rearrange the equation into standard quadratic form for

t

\frac{1}{2}gt^2 - ut + h = 0 \implies t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

The two solutions

t_1

and

t_2

represent the times at which the body reaches the specific height

h

.

3

Apply Vieta's formulas for the sum of roots of the quadratic equation

t_1 + t_2 = \frac{2u}{g}

The sum of the roots of a quadratic equation

t^2 - Bt + C = 0

is equal to the coefficient

B

.

4

Substitute given values to solve for

u

2 + 4 = \frac{2u}{10} \implies 6 = \frac{u}{5} \implies u = 30\text{ m/s}

Given

t_1 = 2\text{ s}

,

t_2 = 4\text{ s}

, and

g = 10\text{ m/s}^2

, direct substitution yields the initial velocity.

Anahtar Kavram

Vertical Motion under Gravity and Time Symmetry

Tahmini Süre:1m 0s

Soru 15Soru →

A bullet of mass $20\text{ g}$ is moving horizontally at a speed of $400\text{ m s}^{-1}$ . It strikes a stationary block of wood and emerges from the opposite side with a speed of $100\text{ m s}^{-1}$ . What is the magnitude of the work done by the bullet against the resistance of the block, in Joules?

Cevabı ve açıklamayı göster

Cevap: 1500

Cevap

The magnitude of the work done by the bullet in penetrating the block is

1500\text{ J}

.

According to the Work-Energy Theorem, the net work done on an object equals the change in its kinetic energy. The reduction in kinetic energy as the bullet slows from

400\text{ m s}^{-1}

to

100\text{ m s}^{-1}

equals the work done against the resistive force of the wooden block.

Adım Adım Çözüm

1

Convert the mass of the bullet into standard SI units (kilograms)

m = \frac{20}{1000} = 0.02\text{ kg}

The standard SI unit for mass in energy equations is the kilogram.

2

Determine the initial kinetic energy of the bullet before entering the block

E_{ki} = \frac{1}{2} \times 0.02\text{ kg} \times (400\text{ m s}^{-1})^2 = 1600\text{ J}

Kinetic energy is defined as

E_k = \frac{1}{2} m v^2

.

3

Determine the final kinetic energy of the bullet as it emerges from the block

E_{kf} = \frac{1}{2} \times 0.02\text{ kg} \times (100\text{ m s}^{-1})^2 = 100\text{ J}

The bullet loses speed upon passing through the block, reducing its kinetic energy.

4

Apply the Work-Energy Theorem to find the work done against resistive forces

W = \Delta E_k = 1600\text{ J} - 100\text{ J} = 1500\text{ J}

The net work done on the bullet equals its change in kinetic energy.

Anahtar Kavram

Work-Energy Theorem

Soru 16Soru →

A simple pendulum of length $l$ has a period of oscillation of $2.0\text{ s}$ when suspended with a bob of mass $50\text{ g}$ . If the bob is replaced with one of mass $200\text{ g}$ while maintaining the exact same string length, what is the new period of oscillation?

2.0\text{ s}

4.0\text{ s}

1.0\text{ s}

8.0\text{ s}

Cevabı ve açıklamayı göster

Cevap: $2.0\text{ s}$

Cevap

The period of oscillation remains

2.0\text{ s}

.

The period of oscillation of a simple pendulum undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{l}{g}}

, where

l

is the length of the pendulum and

g

is the acceleration due to gravity. Because the mass of the bob does not appear in this equation, changing the mass from

50\text{ g}

to

200\text{ g}

does not alter the period. Therefore, the period remains

2.0\text{ s}

.

Adım Adım Çözüm

1

Identify the formula for the period of a simple pendulum.

T = 2\pi \sqrt{\frac{l}{g}}

The period

T

depends on the length of the pendulum

l

and local acceleration due to gravity

g

.

2

Analyze the dependence of period on bob mass.

The mass parameter

m

does not appear in the formula

T = 2\pi \sqrt{\frac{l}{g}}

.

The period of a simple pendulum is independent of the mass of the bob.

3

Determine the new period of oscillation.

T_{new} = 2.0\text{ s}

Since length

l

and acceleration due to gravity

g

remain unchanged, the period stays

2.0\text{ s}

.

Anahtar Kavram

Independence of simple pendulum period from bob mass

Soru 17Soru →

The physical quantity impulse is defined as the product of force and time. Which of the following physical quantities has the same dimensions as impulse?

Linear momentum

Work done

Power

Force

Cevabı ve açıklamayı göster

Cevap: Linear momentum

Cevap

Linear momentum

Linear momentum is calculated as mass times velocity, which yields the base dimensions

[M L T^{-1}]

. This is identical to impulse, which is force times time (

[M L T^{-2}] \times [T] = [M L T^{-1}]

).

Adım Adım Çözüm

1

Determine the dimensions of impulse.

\text{Impulse} = \text{Force} \times \text{Time} = [M L T^{-2}][T] = [M L T^{-1}]

Force has base dimensions

[M L T^{-2}]

and time has dimension

[T]

.

2

Determine the dimensions of linear momentum.

\text{Linear Momentum} = \text{Mass} \times \text{Velocity} = [M][L T^{-1}] = [M L T^{-1}]

Mass has base dimension

[M]

and velocity has dimensions

[L T^{-1}]

.

3

Compare the dimensions of impulse and linear momentum.

Both quantities have identical dimensions of

[M L T^{-1}]

.

By the impulse-momentum theorem, impulse equals change in momentum.

Anahtar Kavram

Dimensional analysis of physical quantities

Soru 18Soru →

A body of mass $0.2\text{ kg}$ undergoes simple harmonic motion with an angular frequency of $10\text{ rad/s}$ and an amplitude of $0.04\text{ m}$ . What is the magnitude of the maximum restoring force acting on the body in newtons?

Cevabı ve açıklamayı göster

Cevap: 0.8

Cevap

The magnitude of the maximum restoring force acting on the body is

0.8\text{ N}

.

The maximum restoring force in simple harmonic motion occurs at maximum displacement (

y = A

) and is given by

F_{\text{max}} = m \omega^2 A

. Substituting

m = 0.2\text{ kg}

,

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

yields

F_{\text{max}} = 0.2 \times (10)^2 \times 0.04 = 0.8\text{ N}

.

Adım Adım Çözüm

1

Identify the given physical quantities from the problem statement.

m = 0.2\text{ kg}

,

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

.

These parameters are required to calculate acceleration and restoring force in simple harmonic motion.

2

Calculate the maximum acceleration of the oscillating body.

a_{\text{max}} = \omega^2 A = (10)^2 \times 0.04 = 4.0\text{ m/s}^2

.

In simple harmonic motion, maximum acceleration occurs at maximum displacement (the amplitude).

3

Determine the maximum restoring force.

F_{\text{max}} = m a_{\text{max}} = 0.2 \times 4.0 = 0.8\text{ N}

.

According to Newton's second law, force is the product of mass and acceleration.

Anahtar Kavram

Maximum Restoring Force in Simple Harmonic Motion

Soru 19Soru →

The mechanical power $P$ dissipated by a circular disc of radius $R$ rotating at an angular velocity $\omega$ in a fluid of density $\rho$ is given by the relation $P = k \cdot \rho^a \cdot \omega^b \cdot R^c$ , where $k$ is a dimensionless constant. Using dimensional analysis, which of the following represents the correct values of the exponents $a$ , $b$ , and $c$ ?

a = 1, b = 3, c = 5

a = 1, b = 2, c = 4

a = 1, b = 3, c = 2

a = 1, b = 1, c = 5

Cevabı ve açıklamayı göster

Cevap: $a = 1, b = 3, c = 5$

Cevap

a = 1, b = 3, c = 5

By writing the dimensions of power as

[M L^2 T^{-3}]

, density as

[M L^{-3}]

, angular velocity as

[T^{-1}]

, and radius as

[L]

, dimensional homogeneity requires that

M L^2 T^{-3} = M^a L^{-3a + c} T^{-b}

. Equating powers gives

a = 1

,

b = 3

, and

c = 5

.

Adım Adım Çözüm

1

Express the base dimensions for each physical quantity

Power

[P] = M L^2 T^{-3}

, Density

[\rho] = M L^{-3}

, Angular velocity

[\omega] = T^{-1}

, and Radius

[R] = L

.

Dimensional analysis requires decomposing derived physical quantities into fundamental dimensions of mass (

M

), length (

L

), and time (

T

).

2

Substitute dimensions into the given equation

P = k \cdot \rho^a \cdot \omega^b \cdot R^c

M L^2 T^{-3} = (M L^{-3})^a \cdot (T^{-1})^b \cdot (L)^c = M^a \cdot L^{-3a + c} \cdot T^{-b}

.

The constant

k

is dimensionless, so its dimension is 1.

3

Equate the powers of

M

,

L

, and

T

on both sides of the equation

For mass

M

:

a = 1

. For time

T

:

-b = -3 \implies b = 3

. For length

L

:

-3a + c = 2 \implies -3(1) + c = 2 \implies c = 5

.

According to the principle of dimensional homogeneity, powers of fundamental dimensions must be equal on both sides of a physically correct equation.

Anahtar Kavram

Dimensional Homogeneity and Formula Derivation

Tahmini Süre:2m 0s

Soru 20Soru →

Match each physical quantity on the left with its correct physical definition and scalar or vector classification on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Work Done matches with Scalar quantity defined as the product of force and displacement in the direction of the force; Acceleration matches with Vector quantity defined as the rate of change of velocity with time; Electric Potential matches with Scalar quantity defined as the work done per unit positive charge in bringing it from infinity; Force matches with Vector quantity defined as the rate of change of linear momentum with time.

Work Done matches the scalar definition involving force and displacement. Acceleration matches the vector definition representing rate of change of velocity. Electric Potential matches the scalar definition of work per unit charge. Force matches the vector definition representing rate of change of momentum.

Adım Adım Çözüm

1

Classify each physical quantity as a scalar (has magnitude only) or a vector (has both magnitude and direction).

Work Done and Electric Potential are scalar quantities. Acceleration and Force are vector quantities.

Scalars require only numerical value and unit, whereas vectors require direction to be fully specified.

2

Match each physical quantity with its precise physical definition.

Work Done corresponds to force multiplied by displacement in the line of action; Acceleration corresponds to velocity change per unit time; Electric Potential corresponds to work done per unit charge; Force corresponds to rate of change of momentum.

Each definition uniquely identifies the fundamental physical relationship for that quantity.

Anahtar Kavram

Classification of physical quantities into scalars and vectors based on their directional properties and definitions

Mechanics

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