Physics question about a bullet?

Question

A 7g bullet moving horizontally at 200 m/s strikes and passes through a 150g tin can sitting on a post. Just after impact, the can has a horizontal speed of 180 cm/s. What was the bullet's speed after leaving the can?

Anybody know?

Answer 1

Conservation of momentum.

mass * velocity of bullet + mass * velocity of can = constant

Mass of bullet: 7 grams = 0.007 kg
Mass of can: 150 grams = 0.150 kg
Initial speed of bullet = 200 m / s
Initial speed of can = 0 m / s

Final speed of can = 1.8 m / s
Final speed of bullet = ??

Initial Energy:
(0.007 kg * 200 m / s) + (0.150 kg * 0 m / s) = 1.4 kg m / sec

Final Energy:
0.007 * bullet final speed + 0.150 kg * 1.80 m / s = 1.4 kg m / sec

0.007 * bullet final speed + 0.27 kg m / s = 1.4 kg m / sec

bullet final speed = (1.4 - 0.27) / 0.007 = 161.4 m / sec

Answer 2

Sure..

Use the conservation of energy principle. That is, the total energy in the system is constant. The energy picked up by the can when the bullet hits it is energy lost by the bullet.

The general expression for kinectic energy is
E = 1/2 * M * V^2

where E=kinectic energy, M = mass, and v = velocity

So,
eb1 = 1/2 * mb (Vb1)^2 where

eb1 = initial energy of the bullet before it hits the can
mb = mass of the bullet and
Vb1 = velocity of the bullet at impact.

The energy that the can picks up as a result of the collision is

Ecan = 1/2 mc * (vc)^2

Ecan = energy of the can
mc = mass of the can
vc = velocity of the can after impact by the bullet.

So the energy of the bullet after impact is the difference between the energy it started with and what was lost to the impact with the can.

eb2 = eb1 - ecan

At this point, you know the energy of the bullet after imact and you know the mass of the bullet.

Using the expression for kinetic energy, substitute the values and solve for V

E = 1/2 * M * V^2 Kinectic energy equation

rearranging,

v = (2E/M) ^0.5

or, vb2 = (2*eb2 / mb)^0.5


where eb2 = eb1 - ecan

You can do the calculations, but that's how you solve the problem.

Hope this helps,

-Guru

Answer 3

this is a classical physics problem, however, there are several assumptions one has to take in order to give a reason answer.

firstly, we asssume the mass of the bullet and the can did not can after the collison (in reality the mass of can would decrease due to the hole made by bullet).
second assumption is that both the can and the bullet travel horizontally after the collision.

Hence using the law of conversation of momentum: momentum before the collision is equal to momentum after the collision. We have
before after
P = P'
P of bullet = P' of Can + P' of bullet.
(bullet) MV = (can) MV + (bullet') MV'
(7*10^-3 * 200) = (150 *10 ^-3 * 1.8) + (7*10^-3 *V' )

solving for V' (bullet speed after) = 161.43 m/s

Note is essential to convert mass and speed in S.I units ie Kg and m/s, respectively.