As shown in the figure below, a bullet is fired at and passes through a piece of

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

(0.536)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.)

Explanation / Answer

Let the system consist of the bullet and the paper.
Since momentum in the system is conserved. Let mass of bullet be m and mass of paper be M.
Let v be initial speed of bullet.

p_f = p_i
M*v_paper + m*0.536v = mv
M*v_paper = 0.464mv
v_paper = [0.464m / M] v............(1)

For conservation of Energy,
dE = transferred energy
KE_f - KE_i = energy transferred
1/2m(0.536v)^2 + 1/2M(v_paper)^2 - 1/2mv^2 = -0.363*1/2mv^2
1/2 * 0.29mv^2 + 1/2M(v_paper)^2 - 1/2mv^2 = -0.363*1/2mv^2
1/2M(v_paper)^2 = 1/2 (1 - 0.29 - 0.363) mv^2
M(v_paper)^2 = 0.347 mv^2
Putting the value of v_paper from above
M[0.464m / M] v]^2 = 0.347 mv^2
M * 0.2153 m^2 v^2 / M^2 = 0.347 mv^2
0.2153 m = 0.347M
M = 0.62 m

Putting this value in (1), we get
v_paper = [0.464 * 0.347 / 0.2153] v
v_paper = 0.748 v

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