Projectile question?

Question

A projectile (mass = 0.191 kg) is fired at and embeds itself in a target (mass = 2.07 kg). (The target is initially stationary). The target (with the projectile in it) flies off after being struck. What percentage of the projectile's incident kinetic energy does the target (with the projectile in it) carry off after being struck?

Answer 1

From the conservation of momentum, we have mV = (m + M)v; so that V/v = (m + M)/m = R the ratio of combined to projectile mass.

Then projectile KE(P) = 1/2 mV^2 and combined masses KE(C) = 1/2 (m + M)v^2 So we are interested in how big one is compared to the other.

To compare similar characteristics take the ratio KE(P)/KE(C) = 1/2 mV^2/1/2 (m + M) v^2 = [m/(m + M)](V/v)^2 = 1/R (R^2) = R; so that KE(P) = KE(C) R = KE(C) [(m + M)/m] or KE(C) = KE(P) (m/(m + M) = KE(P)(.191/2.261) = about 8.4 percent of the projectile's energy is transferred to the combined mass.

The lesson here is that the loss in velocity from V to v more than offsets the gain in mass from m to (m + M); so the result is a loss in kinetic energy. That comes from the fact that KE is proportional to the mass but proportional to the square of the velocity.

Answer 2

If mass of projectile is 'm' and its speed is 'u'

Momentum of projectile 'p'=mu

p^2=m^2u^2-----------------(1)

Kinetic energy E =(1/2)mu^2--------(2)

Dividing (1) by (2)

E = p^2 / 2m--------------------(3)

During this perfectly inelastic collision , momentum 'p' is conserved,the momentum of target and projectile will also be 'p'

Suppose mass of target is 'M' and speed of target with projectile in it is V then, p=(M+m)V


Kinetic energy of target with projectile//Kinetic energy of projectile ={p^2/2 (M+m} / { p^2/2 m}

Kinetic energy of target with projectile /Kinetic energy of projectile = m/ (M+m) = 0.191/ (2.07 +0.191)=0.0844

The target (with the projectile in it) carries off 8.44 % of kinetic energy of projectile