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3 years ago

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There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

l block is 25 kg, and the large block is 50 kg. The blocks are to the left of a hill.The hill is 10 meters tall. (Drawing not to scale)a) The small block collides with the large block and sticks together. How fast must the initial velocityof the small block be so that the two blocks just reach the top of the hill. (Assume no friction).b) What was the total impulse the small block exerted on the large block in order to get it going fastenough

1 answer:

KATRIN_1 [288]3 years ago

6 0

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2 x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75 x 42 / 3

= 1050 Ns.

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$v_c$ is the velocity of the capsule

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$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

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So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

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(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

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For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

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3 years ago