Suppose that one bullet (with mass M = 5:12 g moving to the right at a speed V =

Question

Suppose that one bullet (with mass M = 5:12 g moving to the right at a speed V = 239 m/s directed 21.3 above the horizontal) collides and fuses with another bullet with mass m = 3:05 g moving to the left at a speed v = 282 m/s directed 15.4 above the horizontal.

(a) What is the magnitude (m/s) of their common velocity immediately after the collision?

(b) What is the direction (degrees) of their common velocity immediately after the collision? (Measure this angle from the horizontal.)

(c) What fraction of the original kinetic energy was lost in the collision?

I know this question is similar to this posting

http://www.chegg.com/homework-help/questions-and-answers/suppose-bullet-m-512g-moving-right-speed-250-m-s-directed-213-degrees-horizontal-collides--q2764178

I just can't understand how to simplify and solve the equations the poster provided.

Explanation / Answer

ohk since u already have the solution approach in your hands ... i will explain how to frame the equations to solve this ...
first thing we apply is momentum conservation
momentum = m v
so initial momentum of system = final momentum
system includes both the bullets before and after collision ..
m1v1 + m2v2 = (m1+m2)v3

v1 = 239m/s angle = 21.3

v2 = -282 m/s angle = 15.4

putting the values we have .... v3 = 46.34 m/s angle = 34.8 with horizontal along motion to right ...

Ke lost in collision

=> o.5 (m1+m2)(v3)^2 -[ o.5 (m1)(v1)^2 + o.5 (m2)(v2)^2]

=258.73 KJ

ke fraction lost = 258.73 KJ/[ o.5 (m1)(v1)^2 + o.5 (m2)(v2)^2] = 0.97

or 97 % ke is lost

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