Question

A BB gun is discharged at a cardboard box of mass m2 = 0.65 kg on a frictionless surface. The BB has a mass of m1 = 0.012 kg and flys at a velocity of v1 = 99 m/s. It is observed that the box is moving at a velocity of v2 = 0.39 m/s after the BB passes through it.1. Write an expression for the magnitude of the BB's velocity as it exits the box vf.2. What is the BB's final velocity vf, in meters per second?
3. If the BB doesn't exit the box, what will the velocity of the box v'2, be in meters per second?

Answer 1

Answer:

1. $v_{1f} = v_{1i} - \frac{m_2}{m_1*v_{2f}}$

2. $v_{1f}=77.875m/s$

3. $v_f=1.79m/s$

Explanation:

Use conservation of momentum:

$p_i=p_f$

$m_1*v_{1i} + m_2*v_{2i} = m_1*v_{1f} + m_2*v_{2f}$

1.

The expression for the magnitude of the BB's velocity is

v2i = 0 m/s, solve for v1f

$v_{1f} = v_{1i} - \frac{m_2}{m_1*v_{2f}}$

2.

Now replacing numeric to find the value

$v_{1f} = 99 m/s - \frac{0.65kg}{0.012kg*0.39m/s}$

$v_{1f}=77.875m/s$

3.

Now is doesn't exist the box the velocity will be

v1f = v2f = vf

Conservation of Momentum

$p_i=p_f$

$m_1*v_{1i} = (m_1+m_2)*v_f$

$v_f = m_1*v_{1i} / (m1+m2)$

$v_f = \frac{0.012kg * 99 m/s }{(0.012 kg + 0.65 kg)}$

$v_f=1.79m/s$