A pellet gun is discharged at a cardboard box of mass m_2 = 0.55 kg on a frictio

Question

A pellet gun is discharged at a cardboard box of mass m_2 = 0.55 kg on a frictionless surface. The pellet has a mass of m_ 1 = 0.0185 kg and flys at a velocity of v_1 = 59 m/s. It is observed that the box is moving at a velocity of v_2 = 0.23 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. 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

m2 = 0.55 m1 = 0.0185

V1 = 59 m/s

V2 = 0.23 m/s

momentum conservation

Pi = Pf

m1*v1 = m2*v2 + m1*Vf

Vf = (m1*v1 - m2*v2 ) / m1 .....................answer

now plug in the values

Vf = (0.0185 * 59 - 0.55*0.23) / 0.0185 = 52.16 m/s ............answer

last part

Pi = Pf

m1*v1 = ( m1+m2) *V'

V' = m1*v1 / ( m1+m2

= 0.0185 * 59 / (0.0185 +0.85)

= 1.92 m/s ................answer

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