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 = 0.016 kg and flys at a velocity of v = 87 m/s. It is observed that the box is moving at a velocity of v_2 = 026 m/s after the pellet passes through it. Write an expression for the magnitude of the pellet's velocity as it exits the box v_f. 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

a)

Using conservation of momentum

m1 v1 = m2 v2 + m1 vf

vf = (m1 v1 - m2 v2 ) /m1

b)

vf = (m1 v1 - m2 v2 ) /m1 = (0.016 x 87 - 0.85 x 0.26) /0.016

vf = 73.2 m/s

c)

Using conservation of momentum

m1 v1 = (m2 + m1) v'2

v'2 = m1 v1 /(m2 + m1) = 0.016 x 87 /(0.016 + 0.85) = 1.61 m/s

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