Question

a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. what is their velocity as they roll away together

Answer 1

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$  

where:  

$m_1 = 3 kg$ is the mass of the putty

$u_1 = 10 m/s$ is the initial velocity of the putty (we take its direction as positive direction)

$m_2 = 8 kg$ is the mass of the ball

$u_2 = 0 m/s$ is the initial velocity of the ball (at rest)

$v$ is the final combined velocity of the two putty+ball

Re-arranging the equation and substituting the values, we find the  final combined velocity:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s$

And the positive sign indicates their final direction is the same as the initial direction of the putty.

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