PRINTER VERSIONBACK NEXT Chapter 07, Problem 24 Your answer is partially correct

Question

PRINTER VERSIONBACK NEXT Chapter 07, Problem 24 Your answer is partially correct. Try again. To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.00822-kg bullet is fired straigh up at a falling wooden block that has a mass of 4.90 kg. The bullet has a speed of 571 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t. Number To.0572 the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work Uni

Explanation / Answer

Solution :

Given,

velocity before collision, u1 = 571 j m/s

mass of bullet, m1 = 0.00822 kg

Let velocity of block before collision, u2 = -wj

mass of block, m2 = 4.90 kg

Using law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

4.69 j - 4.90 wj = 4.90822 *wj

4.69 = w*(4.90 + 4.90822)

w = 0.478

But v = u+at

w = 0+gt

t = w/g = 0.478/9.81 = 0.049 sec

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