A 3\text{ kg} object moving at 20\text{ m/s} along a frictionless horizontal surface colli… — Armed Services Vocational Aptitude Battery (ASVAB)

A $3\text{ kg}$ object moving at $20\text{ m/s}$ along a frictionless horizontal surface collides with and sticks to a stationary $7\text{ kg}$ object. Immediately after the collision, the combined mass slides onto a rough surface with a coefficient of kinetic friction of $0.3$ . Assuming the acceleration due to gravity is $10\text{ m/s}^2$ , how many meters will the combined mass slide before coming to a complete stop?

Answer: 6 m

Answer

The problem is solved in two phases. Phase 1 is the perfectly inelastic collision, where momentum is conserved:

p_i = p_f \Rightarrow (3)(20) = (10)v

, which gives a shared velocity of

6\text{ m/s}

. Phase 2 is the sliding on a rough surface, where the only horizontal force is kinetic friction. The acceleration is

a = -\mu_k g = -(0.3)(10) = -3\text{ m/s}^2

. Finally, using the kinematic equation

v_f^2 = v_i^2 + 2ad

(where

v_f = 0

), we get

0 = 36 + 2(-3)d

, yielding

d = 6\text{ meters}

Step-by-Step Solution

Apply the law of conservation of momentum to find the velocity of the combined mass immediately after the collision.

v = 6\text{ m/s}

In a perfectly inelastic collision, momentum is conserved (

m_1v_1 + m_2v_2 = (m_1+m_2)v

). Therefore,

3(20) + 7(0) = 10v

Calculate the deceleration of the combined mass due to kinetic friction.

a = -3\text{ m/s}^2

The net force is kinetic friction (

F_k = \mu_k m g

). Using Newton's second law, acceleration

a = -F_k / m = -\mu_k g = -0.3(10)

Use kinematics to find the distance required for the mass to come to a stop.

d = 6\text{ m}

Using the time-independent kinematic equation

v_f^2 = v_i^2 + 2ad

with

v_f = 0\text{ m/s}

v_i = 6\text{ m/s}

, and

a = -3\text{ m/s}^2

Key Concept

Conservation of Momentum and Kinematics

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic