Mastering Physics Solutions: Exercise 6.46

Mastering Physics Solutions: Exercise 6.46

On January 4, 2012, in Chapter 06: Linear Momentum and Collisions, by Mastering Physics Solutions

C Exercise 6.46

Part A = 0.57m/s Click to use the calculator/solver for this part of the problem

Solutions Below:

A 12 g bullet moving horizontally at 400 m/s penetrates a 2.1 kg wood block resting on a horizontal surface.

Part A

If the bullet slows down to 300 m/s after emerging from the block, what is the speed of the block immediately after the bullet emerges?

Express your answer using two significant figures.

Momentum is given by:

p = mv

And since momentum is conserved, we know the momentum before and after the collision is the same.

Start by solving for beginning momentum (before the collision):

p = mv
p = 0.012 * 400
p = 4.8Ns

Now solve for the velocity:

p = mv
p = mbulletvbullet + mblockvblock
4.8 = (0.012 * 300) + (2.1 * vblock)
4.8 = 3.6 + 2.1vblock
1.2 = 2.1vblock

vblock = 0.57m/s

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