As shown in the figure below, a bullet is fired at and passes through a piece of
ID: 3161989 • Letter: A
Question
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 (0.536)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.433)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=
M=
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An object with a mass of 6.00 g is moving to the right at 14.0 cm/s when it is overtaken by an object with a mass of 26.0 g moving in the same direction with a speed of 22.0 cm/s. If the collision is elastic, determine the speed of each object after the collision.
26.00 g object=
6.00 g object=
Explanation / Answer
According to the given problem,
Given, Post-collision bullet KE = (0.536)² * [KEb BC] = 0.2873 * [KEb BC]
if I understand your nomenclature.
Lost energy = 0.433 * [KEb BC], so the target has
KEt = (1 - 0.433 - 0.2873) * [KEb BC] = 0.28 * [KEb BC]
Since [KEb BC] = ½mv²,
KEt = 0.140mv² = ½MV²
From conservation of momentum we have
mv + 0 = 0.536mv + MV MV = 0.464mv
and M = 0.464mv / V
Plug that into the KEt equation:
0.140mv² = ½(0.464mv / V)V² = 0.233mvV
0.603v = V
and M = 0.464mv / 0.882v = 0.77m
B)
The 1-dimensional classical elastic collision problem is already solved for you:
v1f = (v1i*(m1 - m2) + 2*m2*v2i)/(m1 + m2)
v2f = (v2i*(m2 - m1) + 2*m1*v1i)/(m1 + m2)
Data:
v1i = 14 cm/s
m1 = 6 grams
v2i = 22 cm/s
m2 = 26 grams
Answer for this problem (all are in the same direction as the initial motion for both objects):
A: v1f = 27 cm/s
B: v2f = 19 cm/s
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