Sometimes, in an intense battle, gunfire is so intense that bullets from opposit

Question

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = 228 m/s directed 21.3degree above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4degree above the horizontal. What is the magnitude (m/s) of their common velocity immediately after the collision? What is the direction (degrees) of their common velocity immediately after the collision? (Measure this angle from the horizontal.) What fraction of the original kinetic energy was lost in the collision?

Explanation / Answer

Part A

We know that momentum must be conserved. Furthermore, we know that both the x and y components will be conserved as well.

We will need to break this down into x and y components to solve the problem

Mass 1

Initial Momentum in the x (right) direction = (5.12 X 10-3)(239)(cos 21.3) = 1.14 kgm/s

Initial momentum in the y (up) direction = (5.12 X 10-3)(239)(sin 21.3) = .445 kgm/s

Mass 2

Initial Momentum in the x (left) direction = (3.05 X 10-3)(282)(cos 15.4) = .829 kgm/s

Initial momentum in the y (up) direction = (3.05 X 10-3)(282)(sin 15.4) = .228 kgm/s

So, the total momentum in the x before the collision was 1.14 - .829 = .311 kgm/s

And the total momentum in the y before the collision was .445 + .228 = .673 kgm/s

After the collision, the momenta must remain the same, however the masses fuse, so the combined velocity must be found

i = f

The combined mass is (5.12 X 10-3) + (3.05 X 10-3) = 8.17 X 10-3 kg

In the x,    .311 = (8.17 X 10-3)(vx)

v final x = 38.1 m/s

In the y,    .673 = (8.17 X 10-3)(vy)

v final y = 82.4 m/s

Combining those two velocity vectors using the pythagorean theorem leaves...

c2 = a2 + b2

c2 = 38.12 + 82.42

c = 90.8 m/s

Part B

The direction can be found using right triangle geometry.

tan = O/A

= tan-1 (82.4)/(38.1)

= 65.2o above the horizontal to the right (or North of East, so to speak)

Part C

For this part we need to find the Kinetic Energy relationship

KE = .5mv2

The Initial KE of Bullet 1 = .5(5.12 X 10-3)(239)2 = 146.5 J

The Initial KE of Bullet 2 = .5(3.05 X 10-3)(282)2 = 121.3 J

Total Initial KE = 267.8 J

The final KE of the fused masses is .5(8.17 X 10-3)(90.8)2 = 33.7 J

Percent lost is [(267.8 - 33.7) / 267.8] X 100%

Percent lost is 87.4%

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