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 (0.486)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.363)KEb BC] 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.)

TAC 7r1 (a) Before collision (b) After collision

Answer 1

The momentum of the system bullet-paper must be conserved, so

$mv=MV+mv_{ac}$

$mv=MV+m0.486v$

Solving for M:

$M=\frac{(1-0.486)mv}{V}$

On the other hand, before the collision, the bullet's kinetic energy is

$K_{EbBC}=\frac{1}{2}mv^2$

Then, fro conservation of energy:

$\Delta E=\sum E_{transfered}$

$K_{EbAC}-K_{EbBC}=\sum E_{transferred}$

$\frac{1}{2}m{(0.486v)}^2+\frac{1}{2}M{V}^2-\frac{1}{2}m{v}^2=-0.363\frac{1}{2}mv^2$

$m{(0.486v)}^2+M{V}^2-m{v}^2=-0.363mv^2$

Substituting M:

$m{(0.486v)}^2+\frac{(1-0.486)mv}{V}{V}^2-m{v}^2=-0.363mv^2$

Solving for V:

$V=\frac{(1-0.363-0.236)mv^2}{(1-0.486mv)}$

$V=0.780v$

And the Mass M is:

$M=\frac{(1-0.486)mv}{0.780v}=0.658m$ EDITED