# Math Help - projectile motion, find percentage?

1. ## projectile motion, find percentage?

A projectile (mass = 0.17 kg) is fired at and embeds itself in a target (mass = 2.03 kg).

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?

First off is this in two dimensions? How is the percentage calculated from elastic collision or momentum theorems?

thanks for any help.

2. still haven't figured this one out, anyone?

3. Originally Posted by rcmango
A projectile (mass = 0.17 kg) is fired at and embeds itself in a target (mass = 2.03 kg).

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?

First off is this in two dimensions? How is the percentage calculated from elastic collision or momentum theorems?

thanks for any help.
Once again, there is no net external force on the system during the collision. Let's call the +x direction the direction the bullet travels in. Note that the bullet and block move together as one object after the collision.

So
$P = P_0$

$(M + m)V = mv$
(where M is the mass of the block, m is the mass of the bullet, V is the velocity of the bullet + block, and v is the incident velocity of the bullet.)

$V = \frac{mv}{M + m}$

We want the bullet + block's kinetic energy. Thus
$KE = \frac{1}{2}(M + m)V^2 = \frac{1}{2}(M + m) \frac{m^2v^2}{(M + m)^2}$

$= \frac{m^2v^2}{2(M + m)}$

What fraction of the bullet's incident kinetic energy is this? Well, just before the collision the bullet had a kinetic energy of $\frac{1}{2}mv^2$, so
$\% = \frac{ \frac{m^2v^2}{2(M + m)} }{ \frac{1}{2}mv^2 } \times 100 \%$

$= \frac{m}{M + m} \times 100 \%$

and I'll let you plug the masses in from here.

Note that this result says there is no way to make this collision elastic.

-Dan

4. okay, i got 7.7 %.
thankyou very much for the supplied work w/ explanation.