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Difficulty: HardNewton's Laws of Motion and Linear Momentum

A body of mass 3.0 kg3.0\text{ kg} moving due east along a smooth horizontal track at 8.0 m s−18.0\text{ m s}^{-1} collides head-on with a 2.0 kg2.0\text{ kg} body moving due west at 4.0 m s−14.0\text{ m s}^{-1}. Immediately after the impact, the 2.0 kg2.0\text{ kg} body rebounds due east with a speed of 5.0 m s−15.0\text{ m s}^{-1}. If the duration of the impact is 0.02 s0.02\text{ s}, what is the magnitude of the average impact force exerted on the 3.0 kg3.0\text{ kg} body?

  1. A
    100 N100\text{ N}
  2. B
    1200 N1200\text{ N}
  3. 900 N900\text{ N}Answer
  4. D
    1500 N1500\text{ N}

Answer

The magnitude of the average impact force exerted on the 3.0 kg3.0\text{ kg} body is 900 N900\text{ N}.
By designating East as positive, initial velocities are u1=+8.0 m s−1u_1 = +8.0\text{ m s}^{-1} and u2=−4.0 m s−1u_2 = -4.0\text{ m s}^{-1}. Conservation of linear momentum gives (3.0×8.0)+(2.0×−4.0)=3.0v1+(2.0×5.0)(3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0), yielding v1=+2.0 m s−1v_1 = +2.0\text{ m s}^{-1}. The change in momentum of the 3.0 kg3.0\text{ kg} body is Δp=3.0×(2.0−8.0)=−18.0 N s\Delta p = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}. Dividing the magnitude of this impulse by the impact time (0.02 s0.02\text{ s}) yields an average force of 900 N900\text{ N}.

Step-by-Step Solution

1
Set up momentum conservation by assigning directional signs to velocities.
Taking East as positive (++): u1=+8.0 m s−1u_1 = +8.0\text{ m s}^{-1}, u2=−4.0 m s−1u_2 = -4.0\text{ m s}^{-1}, v2=+5.0 m s−1v_2 = +5.0\text{ m s}^{-1}.
Linear momentum is a vector quantity, so direction must be taken into account.
2
Calculate the initial total momentum and solve for the final velocity of the 3.0 kg3.0\text{ kg} body (v1v_1).
m1u1+m2u2=m1v1+m2v2  ⟹  (3.0×8.0)+(2.0×−4.0)=3.0v1+(2.0×5.0)  ⟹  16.0=3.0v1+10.0  ⟹  v1=+2.0 m s−1m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies (3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0) \implies 16.0 = 3.0 v_1 + 10.0 \implies v_1 = +2.0\text{ m s}^{-1}.
Total momentum is conserved in the absence of external forces.
3
Calculate the magnitude of the impulse and average force acting on the 3.0 kg3.0\text{ kg} body.
Δp1=m1(v1−u1)=3.0×(2.0−8.0)=−18.0 N s\Delta p_1 = m_1(v_1 - u_1) = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}. Force magnitude F=∣Δp1∣Δt=18.00.02=900 NF = \frac{|\Delta p_1|}{\Delta t} = \frac{18.0}{0.02} = 900\text{ N}.
The average force equals the rate of change of linear momentum.

Key Concept

Law of Conservation of Linear Momentum and Impulse-Momentum Theorem
Estimated Time:2m 0s
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Topics

Newton's Laws of Motion and Linear Momentum

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