Q1 to 2! y to show that the fractional loss in kinetic Using equations (18.11) a
ID: 1783119 • Letter: Q
Question
Q1 to 2!
y to show that the fractional loss in kinetic Using equations (18.11) and (18.13) and equations (18.6) and (18.8), it is eas energy for the "ball-pendulum" system is given by 7. (18.14) m+M Use equation (18.14) and find the fractional loss in kinetic energy. Record the result in Data T fractional loss with the fractional loss calculated in step 6 by computing the percent difference. Data Table 18.3. Make conclusions. able 18.3. Compare this Record your results in Questions A 15 g bullet is fired horizontally into a block of wood with a mass of 2.5 kg and embedded in the block. Initially the block of wood hangs vertically and the impact causes the block to swing so that its center of mass rises 15 cm. Find the velocity of the bullet just before the impact. Prove that the fractional loss in kinetic energy for the "ball-pendulum" system is given by equation (18.14). 1. 2.Explanation / Answer
1)
15 g bullet is fired horizontally into a block of wood with a mass of 2.5kg and embedded in the block. Initially the block of wood hangs vertically and the impact causes the block to swing so that its center of mass rises 15 cm. Find the velocity of the bullet just before the impact.
Initial horizontal momentum = 0.015 * vi
Use conservation of momentum to determine the velocity of the block with bullet embedded. Since the bullet is embedded in the block, the final momentum = total mass * velocity
Total mass = 0.015 kg + 2.5 kg = 2.515 kg
Final horizontal momentum = 2.515 * vf
2.515 * vf = 0.015 * vi
vf = (0.015 ÷ 2.515) * vi
This is the velocity of the block after the bullet was embedded in it.
Use conservation of energy to determine the velocity of the bullet before it hits the block. As the block with embedded bullet rises, its kinetic energy decreases and its potential energy increases.
Initial KE = Final PE
Initial KE = ½ * 2.515 * [(0.015 ÷ 2.515) * vi]^2
Final PE = 2.515 * 9.8 * 0.15 = 3.69705
Set KE = PE and solve for vi
½ * 2.515 * [(0.015 ÷ 2.515) * vi]^2 = 3.69705
(½ * 2.515 * 0.015^2 ÷ 2.515^2) * vi^2 = 3.69705
(½ * 0.015^2 ÷ 2.515) * vi^2 = 3.69705
vi^2 = 3.69705 ÷ (½ * 0.015^2 ÷ 2.515)
vi = 287.5 m/s (ans)
Let’s Check:
vf = (0.015 ÷ 2.515) * vi
vf = (0.015 ÷ 2.515) * 287.5
KE = ½ * 2.515 * [(0.015 ÷ 2.515) * 287.5] ^2 = 3.69705
The KE is equal to the PE, so the answer is correct!
I hope this helps you understand how to solve this type of problem.
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