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2 years ago

10

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

ultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide

1 answer:

Nonamiya [84]2 years ago

8 0

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

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So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

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$\Delta t = 0.600 s$

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:

$K=\frac{1}{2}mv^2$

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$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

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

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