A 6.0 gram bullet trawling at 2Sd m s enters and stops in a 2.10 kg block of woo
ID: 1418723 • Letter: A
Question
A 6.0 gram bullet trawling at 2Sd m s enters and stops in a 2.10 kg block of wood. What is the speed of the block/bullet combination? What fraction of the system's original Kinetic energy was in the perfectly inelastic collision? How much work was done in stopping the bullet? If the bullet traveled 5 cm into the wood before stopping, What was the (assumed constant) force acting on the bullet causing it to stop? A 2.5 kg cart traveling at 26 cm/s collides elastically (and head-on) with a 1.6 kg cart initially at rest. What are the final velocities of the two carts? Assume a perfectly elastic collision between a 0.916 kg baseball bat and a 0.145 kg baseball traveling in the opposite direction. The speed of the ball was 32 m/s before the collision and 40 m/s after the collision. What was the speed of the bat? Two skaters (with masses 65 kg and 85 kg) initially at rest push away from each other. After the push, the lighter skater's speed is 2.35 m/s. How fat is the heavier skater moving? An explosion can be thought of as a time-reversed perfectly inelastic collision. As As one example, consider a neutron at rest that decays into a proton and an electron. Ejected in opposite directions. If the speed of the electron is 7.25 Times 10^9 m/s, what is the speed of the proton?Explanation / Answer
At a single time, I can solve maximum 4 questions.
9) m1 = 2.5, m2 = 1.6, v1 = 0.26 m/s, v2 = 0
Now V1 = [v1(m1 – m2) + 2m2v2]/(m1 + m2) = 5.7 cm/s
And V2 = [v2(m2 – m1) + (2m1v1)]/(m1 + m2) = 31.7 cm/s
10) m1 = 0.916 kg, m2 = 0.145 kg, v2 = 32 m/s, V2 = 10 m/s, v1 = ?
Now V2 = [v2(m2 – m1) + (2m1v1)]/(m1 + m2)
From here v1 = 19.26 m/s
11) m1 = 65 kg, m2 = 85 kg, v1 = v2 = 0, V1 = 2.35 m/s, V2 = ?
Momentum conservation, m1v1 + m2v2 = m1V1 + m2V2
0 = m1V1 + m2V2 ==========> V2 = -1.8 m/s
12) Momentum conservation,
0 = 9.1*10^-31*7.25*10^6 + 1.67*10^-27*V
V = -3950.6 m/s
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