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svlad2 [7]
3 years ago
10

Two blocks with a mass of 2 kg collide in an elastic collision. The initial velocity of Block 1 is 10 m/s. Block 2 is traveling

at 2 m/s in the same direction as Block 1. What is the final momentum and kinetic energy of Block 2?
A final momentum: 2 kg x m/s, final kinetic energy: 1 J
B. final momentum: 20 kg x m/s, final kinetic energy: 10 J
C final momentum: 90 kg x m/s, final kinetic energy: 20 J
D. final momentum: 22 kg x m/s, final kinetic energy: 101 J​
Physics
1 answer:
pentagon [3]3 years ago
4 0

Answer:

p = 20 kg•m/s

KE = 100 J

Explanation:

In an elastic collision of identical masses, the two masses will exchange momentums. Therefore Block 1 initially moving at 10 m/s will be moving at 2 m/s, and Block 2 will go from 2 m/s to 10 m/s

momentum = mv = 2(10) = 20 kg•m/s

KE = ½mv² = ½(2)10² = 100 J

Unfortunately, your answer selection does not have this answer as an option.

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A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
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e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

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<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

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p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

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<em>& </em><u><em>Impulse:</em></u>

I=F.dt

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I=m.\frac{dv}{dt}.dt

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