Answered

A 0.0010-kg pellet is fired at a speed of 50.0m/s at a motionless 0.35-kg piece of balsa wood. When the 

pellet hits the wood, it sticks in the wood and they slide off together. With what speed do they slide?

Verified Answer :

p = m*v

conservation of momentum suggests

initial momentum equals final momentum

mv-initial = mv-final

(0.0010 kg)(50 m/s) = (0.0010 kg + 0.35 kg)v

thus:

v = (0.0010)(50)/(0.351) = 0.142 m/s

Answer: The pellet and wood slide at a speed of 0.142 m/s.

Explanation:

To calculate the velocity of the pellet and wood after the collision, we use the equation of law of conservation of momentum, which is:

\(m_1u_1+m_2u_2=m_1v_1+m_2v_2\)

where,

\(m_1\) = mass of pellet = 0.001kg

\(u_1\) = Initial velocity of pellet = 50m/s

\(v_1\) = Final velocity of pellet = v m/s

\(m_2\) = mass of wood = 0.36kg

\(u_2\) = Initial velocity of wood = 0m/s

\(v_2\) = Final velocity of wood = v m/s

Putting values in above equation, we get:

\((0.001\times 50)+(0.35\times 0)=(0.001+0.35)v\\\\v=0.142m/s\)

Hence, the pellet and wood slide at a speed of 0.142 m/s.

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