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lidiya [134]
3 years ago
12

A(n) 0.5 kg soccer ball approaches a player horizontally with a speed of 10.6 m/s. The player illegally strikes the ball with he

r hand and causes it to move in the opposite direction with a speed of 19.9 m/s. What is the magnitude of the impulse delivered to the ball by the player? Answer in units of kg · m/s.
Physics
1 answer:
Llana [10]3 years ago
6 0

Answer: - 25.2 kgm/s

Explanation: The mass of the ball is 0.5kg, and the initial velocity = 10.6m/s.

The final velocity is in opposite direction of the initial hence final velocity (v) = - 19.9 m/s

Impulse = change in momentum = final momentum - initial momentum.

Final momentum = mass × final velocity

Final momentum = - 19.9 × 0.5

Final momentum = - 9.95 kgm/s

Initial momentum = mass × initial velocity

Initial momentum = 0.5 × 10.6 = 5.3kgm/s

Change in momentum = final momentum - initial momentum = - 19.9 - 5.3

Change in momentum = - 25.2 kgm/s

The negative sign implies that the change in momentum is the opposite direction relative to the first.

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3 years ago
Please help! I'm not sure what equation or the process to do this question.
lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

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And the kinetic energy of the ball is

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When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

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The 3rd one. The question can be tested by a systematic procedure
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