A pellet gun is shot at a cardboard box of mass m_2 = 0.85 kg on a frictionless

Question

A pellet gun is shot at a cardboard box of mass m_2 = 0.85 kg on a frictionless surface. The pellet has a mass of m_1 = 0.0175 kg and moves at a velocity of V_1 = 96 m/s. It is observed that the box is moving at a velocity of v_2 = 0.48 m s after the pellet goes through it. Write an expression for the magnitude of the pellet's velocity as it exits the box v_f. What is the pellet's final velocity V_f, in meters per second? If the pellet doesn't exit the box. what will the velocity of the box v'2, be in meters per second?

Explanation / Answer

Momentum is always conserved. Since the box is initially at rest, the final momentum of the bullet is equal to its initial momentum minus the final momentum of the box.

0.0175 * vf = 0.0175 * 96 – 0.85 * 0.48
0.0175 * vf = 1.272
vf = 1.272 / 0.0175

vf = 72.69

This is approximately 72.69 m/s .If the bullet does not exit the box, they will have the same velocity after the collision.

now v1f = v2f = vf
Back to the Conservation of Momentum
m1*v1i = (m1+m2)*vf
vf = m1*v1i / (m1+m2)
vf = 0.0175 kg * 96 m/s / (0.0175 kg + 0.85 kg) = 1.94 m/s

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