Question

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.496)v after passing through the target. The collision is inelastic and during the collision, the amount of kinetic energy lost by the bullet and paper is equal to [(0.263)Kb BC] , that is, 0.263 of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)

V =  ? v

M =  ? m

Answer 1

We can solve this problem using conservation of momentum and conservation of kinetic energy.

Using Conservation of momentum

Let mass of the bullete = m and velocity of the bullet = v

Mass of the paper = M and velocity of the Paper = V

Momentum before collision =

p_{before} = mv

Momentum after collision

p_{After} = m*0.496v + MV

p_{before} = p_{After}

mv = m*0.496v + MV   --- equation (1)

Using conservation of energy

Kinetic energy before collision

KE_{before} = 1/2 mv^2

Kinetic energy before collision

KE_{after} = 1/2 m(0.496v)^2 + 1/2MV^2 + 1/2mv^2*0.263

KE_{before} = KE_{after}

1/2 mv^2 = 1/2 m(0.496v)^2 + 1/2MV^2 + 1/2mv^2*0.263  

mv^2 - m(0.496v)^2 - mv^2*0.263 = MV^2

mv^2[1-0.246 - 0.263] = MV^2           

mv^2*0.491 = MV^2 ---------------- equation (2)

Dividing equation 2 with 1 we get.

v*0.974 = V

using this value in eqt 1 we get

m*0.517 = M

So

V = (0.974)v

M = (0.517)m