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A ball of mass 0.6kg, initially at rest; acquires speed of 4.7m/s immediately after being kicked by force of strength 21N: For how long did this force act on the ba...

Question

A ball of mass 0.6kg, initially at rest; acquires speed of 4.7m/s immediately after being kicked by force of strength 21N: For how long did this force act on the ball?

A ball of mass 0.6kg, initially at rest; acquires speed of 4.7m/s immediately after being kicked by force of strength 21N: For how long did this force act on the ball?



Answers

A ball of mass 0.5 $\mathrm{kg}$ , initially at rest, acquires a speed of 4 $\mathrm{m} / \mathrm{s}$ immediately after being kicked by a force of strength 20 $\mathrm{N} .$ For how long did this force act on the ball?
(A) 0.01 $\mathrm{s}$
(B) 0.1 $\mathrm{s}$
(C) 0.2 $\mathrm{s}$
(D) 1 $\mathrm{s}$

So in this problem, we're kicking a ball. So this is the foot visible and the mass of the bull is ah, 0.5 kilograms. And ah, After kicking the bull, the bull has a velocity vehicle full meters per second. And we know that on average force for the process f it's 20 notice. So the count that time for, uh uh the content time is that tea? We have times t equal m times meat. Right? So we were turned at the T ea quo and B diva f. So this gives you 0.1 seconds.

Hi first question told the Ebola of mass. I am the court 3.4-5 telegrams And was initially arrest to the initial Velocity one. So so everyone is in trouble lost. Tv to finally lost. T. M. Is the mass to use the time F. S. D. Force and mm was killed by one of the players brahmos mom. The final velocity We trained six m/s and the time it's milliseconds and merely is Time strength and power three mainly. This time stand power negative three. So this it's times trans problem one is 3 seconds. Malawi get difficult to 0.4 25 Bracket 2600 Over eight times turned negative three. And this force is to go to 1381 Newtons. So that's our final answer.

Okay, so you exert a force involved when you toss it a past. That is correct. So how long will that first last? All right, so that's the question. So as you apply some force, the force will give the give the ball acceleration on that will just be will that if in Dublin, be finished as soon as the ball, As soon as you lose the boat as soon as the ball leaves her hand. So the answer would be on the first last deal the ball is in contact with, huh? All right. It stops as soon as you really is the ball case of this fence. The answer to this question.

For this problem. On the topic of momentum and collisions, We are told that a ball which has a massive 0.265 kgs, is initially addressed. It is then kicked at an angle of 20.8° to the horizontal, and travels a horizontal distance of 52.8 m after the kick, we want to know the impulse that is received by the ball during this cake. Now we'll use the definition of impulse as the change in momentum, with the initial momentum being zero, since the ball is initially addressed to determine the initial speed or the speed of the ball after being kicked. And then we can apply an equation of motion to get the maximum height. So we know that the impulse jay is equal to delta P, which is just um V not since the final velocity is zero, the initial velocity of the ball is zero. Now, in order to find the speed V, we know that the time for the ball to travel the distance D is equal to two times be not why over G. From the equations of motion, which is two times the initial speed V not times science data which gives us the Y component divided by G. And the distance D. We know is also we not X times TD, which is we not co signed E to the X component of the velocity times the time from above. To be not sign data over G, which gives us Vinod squared sign to the to divided by G. Which means that the speed at which the ball is kicked. The note is equal to the square root of its horizontal distance. D. Times G over sign of two times the angle which is kicked to the to so therefore the impulse G is simply the mass times this velocity. So it's the mass times the square root of the times G Times The Sine of two Theta. We know all of these values. So we can calculate this impulse. The message of the ball is 0.265 Kg. This is multiplied by the square root of its horizontal distance D, which is 52 0.8 m Times G, 9.81 m a square second, divided by the sign of two times the kick angle, which is two times 20.8°.. So if we calculate this, we get the impulse exerted on the soccer ball to be 7.4 kg, meet up for a second.


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