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A 0.2-kg ball moving at 2.0 m/s perpendicular to a wall rebounds from the wall at 1.5 m/s. The impulse given to the ball is:

zero 3.5 N-s away from wall 3
.5 N.s toward wall
0.7 N-s away from wall
0.7 N-s toward wall

Respuesta :

Answer:

J = 0.7 N-s away from wall

Explanation:

Given that,

Mass of the ball, m = 0.2 kg

Initial speed of the ball, u = 2 m/s

Final speed of the ball, v = -1.5 m/s (as it rebounds)

Impulse is equal to the change in momentum. So,

[tex]J=m(v-u)\\\\J=0.2\times (-1.5-2)\\\\J=-0.7\ N-m[/tex]

So, the impulse given to the ball is 0.7 N-m and it is away from the wall. Hence, the correct option is (c).

The impulse given to the ball is J = 0.7 N-s away from the wall.

Calculation of the impulse:

Since

Mass of the ball, m = 0.2 kg

Initial speed of the ball, u = 2 m/s

Final speed of the ball, v = -1.5 m/s (as it rebounds)

So, here the impulse should be

J = 0.2*(-1.5-2)

= -0.7 N-s

Therefore, The impulse given to the ball is J = 0.7 N-s away from the wall.

Learn more about impulse here: https://brainly.com/question/16324614

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