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Note: I\'m in an algebra-based physics class so please show those types of equat

ID: 1332654 • Letter: N

Question

Note: I'm in an algebra-based physics class so please show those types of equations and how to get to the answer.

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.546)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.443)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)

V = v M = m

Explanation / Answer

post-collision bullet KE = (0.546)^2 * [KEb BC] = 0.298*[KEb BC]
by your nomenclature.
Lost energy = 0.443 * [KEb BC], so the target has
KEt = (1 - 0.443 - 0.298) * [KEb BC] = 0.259 * [KEb BC]
Since [KEb BC] = 0.5*mv^2,
KEt = 0.129mv^2 = 0.5MV^2

From conservation of momentum we have
mv + 0 = 0.546mv + MV

MV = 0.454mv
and M = 0.454mv / V
Plug that into the KEt equation:
0.129mv^2 = 0.5(0.454mv / V)V^2

0.129*v = 0.227V

V = 0.129v/0.227 = 0.568v

V = 0.568v
and M = 0.454mv / 0.568v = 0.799m

M = 0.799m

comment below if you have any further doubt

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