PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<
PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<
1st Edition
ISBN: 9781323834831
Author: Knight
Publisher: PEARSON C
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Chapter 11, Problem 1CQ
To determine

Momentum of mass blocks from largest to smallest.

Expert Solution & Answer
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Answer to Problem 1CQ

Solution:The desired order of momenta is (px)b>(px)a=(px)c=(px)e>(px)d .

Explanation of Solution

Formula used:

Momenta (p) = mass of the object (m) × velocity of the object (v)

Given:

Different mass blocks moving with different velocities are given. Just by multiplying the mass of the object with their velocities will give the momenta of the object.

  PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<, Chapter 11, Problem 1CQ

  Figure.1

Calculation:

So, let’s first calculate the momentum of each block.

Momenta of block ‘a’ (px)a=20g×1m/s=20g-m/s

Momenta of block ‘b’ (px)b=20g×2m/s=40g-m/s

Momenta of block ‘c’ (px)c=10g×2m/s=20g-m/s

Momenta of block ‘d’ (px)d=10g×1m/s=10g-m/s

Momenta of block ‘e’ (px)e=200g×0.1m/s=20g-m/s

In the above calculation, we find that the block b has the highest momentum, block a, c and e have the same momenta that is 20g-m/s and block d has the smallest momenta.

Conclusion:

By the calculation, momenta for each block has been calculated and it has been arranged in the order from largest to smallest as follows.

  (px)b>(px)a=(px)c=(px)e>(px)d .

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Chapter 11 Solutions

PHYS 212 FOR SCI+ENG W/MAST PHYS >ICP<

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