A BB gun is fired at a cardboard box of mass m2=0.55 kg on a frictionless surfac

Question

A BB gun is fired at a cardboard box of mass m2=0.55 kg on a frictionless surface. The BB has a mass of m1= 0.0195kg and travels at a velocity of v1= 74 m/s. It is observed that the box is moving at a velocity of v2= 0.21 m/s after the BB passes through it.

A.) Write an expression for the magnitude of the BB's velocity as it exits the box vf.

B.) What is the BB's final velocity vf, in meters per second?

C.) If the BB doesn't exit the box, what will the velocity of the box v'2, be in meters per second?

Explanation / Answer

Given

   m1 mass of BB = 0.0195 kg

   m2 mass of box = 0.55 kg

  
u1 velocity of BB before collision = 74 m/s

v2 is velocity of box after the BB passes through it = 0.21 m/s

by conservation of momentum , and assume that the box is initially at rest

A)

the momentumof the system beforce colllision = momentum of the system after the collision

   m1 u1 + m2* u2 = m1v1 + m2v2

here v1 is BB's final velocity , when it exits the box

   v1 = (m1u1 - m2v2)/m1

B) V1 = (0.0195*74 - 0.55*0.21)/0.0195 m/s

   = 68.1 m/s

C)

   if BB does not exit

then conservation of energy is

   m1*u1 = (m1+m2)V

   V = m1*u1/(m1+m2)
   = 0.0195*74/(0.0195+0.55) m/s

   = 2.53380 m/s

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