CONSERVATION OF MOMENTUM PRACTICE PROBLEMS 2. A 5.0 g bullet Is shot into a 3.0

Question

CONSERVATION OF MOMENTUM PRACTICE PROBLEMS

2. A 5.0 g bullet Is shot into a 3.0 kg wooden block standing on a frictionless sulfate

a) The block, with the bullet in it, acquires a velocity ot 1 .0 m/s. Calculate the velocity ot the bullet before striking the block.

b)IF the bullet were initially traveling at 315 m/s and ricochets otl the block with a speed 250 m/s, what is the velocity of the wooden block after the bullets hits it?

C) if the bullet from part "b" travels through the block and, after going all the way through the block has a speed Of 250 m/s, what is the final velocity of the block?

2. A 2.0 kg object, A, moving at velocity 4.0 m/s, collides with another 5.0 kg

Explanation / Answer

3) mass of rocket, m1 = 5600 kg

velocity of rocket, v1 = 150 m.s

mass of gases expelled, m2 = m (say)

velocity of gases expelled, v2 = 4200 m/s(perpendicular to v1)

let direction of motion of rocket be X-axis and direction of gases expelled by Y

So Initial momentum in X - axis = 5600 * 150 = 840000 kg - m/s

Initial momentum in Y-axis = 0

After expelling the gases, we get

Final Momentum in X - axis = (5600-m) * v * cos40 = 840000 (Initial momentum in X-axis)

Final momentum in Y-axis = - (5600 - m) * v * sin40 + m * 4200 = 0 (Initial momentum in Y -axis)

let 5600 - m be x

so vx = 840000 / cos40, substitute this in the secodn equation, it becomes:

- 840000 * (sin40/cos40) + m * 4200 = 0, so we get m = 167.82 kg

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.