Question

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?

Answer 1

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 2

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

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