As shown in the figure below, a bullet is fired at and passes through a piece of
ID: 2240392 • 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
after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction
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.)
Explanation / Answer
and mass of paper be M. Let v be initial speed of bullet.
p_f = p_i
M*v_paper + m*0.506v = mv
For conservation of Energy,
?E = ?transferred energy
KE_f - KE_i = energy transferred
1/2m(0.506v)^2 + 1/2M(v_paper)^2 - 1/2mv^2 = -0.293*1/2mv^2
solve and get the answers ...
M*v_paper + m*0.506v = mv
M*v_paper = mv - m*0.506v = mv (1-0.506) = mv 0.494 eq 1
using eq1 in below equation:
1/2 m(0.506v)^2 + 1/2 M(v_paper)^2 - 1/2 mv^2 = -0.293*1/2 mv^2
1/2 m(0.506v)^2 + 1/2mv 0.494 (v_paper) - 1/2mv^2 = -0.293*1/2 mv^2
(0.506v)^2 + v 0.494 (v_paper) - v^2 = -0.293* v^2
(0.506^2)v + 0.494 (v_paper) - v = -0.293* v
0.494 (v_paper) = -0.293* v - (0.506^2)v +v
0.494 (v_paper) = v {1-0.256 - 0.293} = 0.451 v
V / v_paper = 0.913 v
M*v_paper = mv 0.494
M*0.913 v = mv 0.494
M*0.913 = m 0.494
M = 0.541 m
final answers :
M = 0.541 m
V / v_paper = 0.913 v
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