Question

A 23. 6 g ball of clay is thrown horizontally to the right at 40. 4 m/s toward a 1. 70 kg block sitting at rest on a frictionless surface. The clay hits and sticks to the block.


What is the speed of the block and clay right after the collision?


Answer in m/s to three significant figures

Answer 1

Answer:

0.142 m/s

Explanation:

You solve it by conservation of momentum.

Before the impact the total momentum is, considering right the positive direction,

$m_1v_1+m_2v_2 =6\cdot10^-^3\times 40.4\ kg\cdot m/s + 1.7\times 0 \ kg\cdot m/s =242.4\cdot 10^-^3\ kg\cdot m/s$

After the impact, we have a single body of mass $m_1+m_2$ and thus a single value of speed, V.

$(m_1+m_2)V =1.703 kg \cdot V$

Since no external forces are applied, the two quantities are equal after the impact. That allows us to solve for V

$V= (242.4\cdot10^-^3 \ kg\cdot m/s )\div (1.703\ kg)=142.337\cdot10^-^3 m/s = 0.142 m/s$