A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. Immediately after the collision, the 2.0-kg object has a velocity of 1.5 m/s directed 30° from its initial direction of motion. What is the x component of the velocity of the 1.0-kg object just after the collision?
A) 3.7 m/s
B) 3.4 m/s
C) 1.5 m/s
D) 2.4 m/s
E) 4.1 m/s

Question

Quizplus · Accepted Answer

To solve this problem, we use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision. The x-component of the momentum for the system before the collision is given by the momentum of the 2.0-kg object, since the 1.0-kg object is at rest. The momentum of the 2.0-kg object is $2.0 \, \text{kg} \times 3.0 \, \text{m/s} = 6.0 \, \text{kg} \cdot \text{m/s}$.After the collision, the x-component of the momentum of the 2.0-kg object is $2.0 \, \text{kg} \times 1.5 \, \text{m/s} \times \cos(30^\circ)$. The cosine of 30 degrees is $\sqrt{3}/2$, so the x-component of the momentum for the 2.0-kg object after the collision is $2.0 \times 1.5 \times \sqrt{3}/2 = 1.5\sqrt{3} \, \text{kg} \cdot \text{m/s}$.Let $v_x$ be the x-component of the velocity of the 1.0-kg object after the collision. The total x-component of the momentum after the collision is $1.5\sqrt{3} + 1.0 \times v_x$.Setting the total momentum before equal to the total momentum after, we get:$$6.0 = 1.5\sqrt{3} + v_x$$Solving for $v_x$, we find:$$v_x = 6.0 - 1.5\sqrt{3}$$$$v_x \approx 6.0 - 1.5 \times 1.732$$$$v_x \approx 6.0 - 2.598 \approx 3.4 \, \text{m/s}$$Therefore, the x-component of the velocity of the 1.0-kg object just after the collision is approximately 3.4 m/s.

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