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 

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:

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

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