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Answer :
To solve for the ball's velocity when the bullet bounces backward, the conservation of momentum equation is used, considering the initial and final velocities and masses of both the bullet and the ball. The ball is found to be moving at approximately 5.33 m/s.
To calculate how fast the 3.48-kg steel ball is moving when the 38.8-g bullet bounces backward at a speed of 5.17 m/s, we apply the principle of conservation of momentum. The initial momentum of the system (bullet plus ball) must equal the final momentum since no external forces are involved. The initial momentum is only due to the moving bullet, as the ball is at rest.
The formula for conservation of momentum is: m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final, where m1 and m2 are the masses of the bullet and the ball, respectively, and v1 and v2 are their velocities.
Given:
m1 (bullet) = 0.0388 kg
v1_initial (bullet) = 472 m/s
v1_final (bullet) = -5.17 m/s (negative since it bounces back)
m2 (ball) = 3.48 kg
v2_initial (ball) = 0 m/s (at rest)
To find:
v2_final (ball's velocity)
Using the conservation of momentum:
0.0388 kg * 472 m/s + 3.48 kg * 0 m/s = 0.0388 kg * (-5.17 m/s) + 3.48 kg * v2_final
Let's solve for v2_final:
18.3136 kg m/s = -0.200556 kg m/s + 3.48 kg * v2_final
v2_final = (18.3136 kg m/s + 0.200556 kg m/s) / 3.48 kg
v2_final ≈ 5.33 m/s
Therefore, the ball is moving at approximately 5.33 m/s when the bullet bounces backwards
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