questions-2022-refer-to-the-following-scenario-a-2-mathrmkg-ball-with-a-velocity-of-20-mathrmm-mat-2

Question

Questions 20–22 refer to the following scenario:
A 2 $\mathrm{kg}$ ball with a velocity of $-20 \mathrm{m} / \mathrm{s}$ collides with the wall and bounces back with a velocity of 10 $\mathrm{m} / \mathrm{s}$
If the ball is in contact with the wall for 0.002 s, determine the average force experienced by the ball.
(A) $3,000 \mathrm{N}$
(B) $5,000 \mathrm{N}$
(C) $30,000 \mathrm{N}$
(D) $50,000 \mathrm{N}$

Sachin Rao · Accepted Answer

in this problem, we have to find the average force experienced by the ball. So here we will use impulse momentum theory. We know that f into Delta T is equal to M into Delta V. So this is the average force is equal to mass into. Delta is the final minus, the initial divided by Delta T. So this is equal to two and TV finalists. 10 and we initially is minus 20 and time is 0.2 seconds. So now we will solve this. This is equal to 30,000 Newton. So this is the average force experienced by the ball during the collision. So now we will check our options. So options see is correct. Often see is correct. Yeah, This is the answer. I hope I answered your question. Thank you.

Salamat Ali · Answer

the average force experience by the ball when it hits the wall is force physical to rate of change of momentum. Dp derided by Delta T uh, where we have a DP From above question, we have 60 new two seconds divided by the time given. We have a 0.0. There are two seconds. So the writing these two numbers gives us the force over 30,000 Newton 30,000 You didn&#x27;t, So her option see is the right option.

Salamat Ali · Answer

The work done by the wall on the ball is given by the view is the call toe right off? Change of frantic energy. Not right off, Jane. Sorry. Changing Patrick Energy. The kind of energy change me. Harry is one of two turned the Maas final Velocity square a minus initial loss. T square, eso substituting. They&#x27;re aliens. One over to where mass we have is a two kg final lost in the heavy areas. There. 10. So square minus 20 square usually lost. So 20 square. This gives us the energy. So he worked on by the wall. On the ball is a minus 300 jewels, 300 jewels. So option is the right answer.

Manish Kumar · Answer

so basically abolish, you know, really contracted with the vault for sometimes. So we know that the impulse is nothing but after 30 over there will be changing moment. Um, that means force exerted by the wall on the bodies. And obviously there is a M v final manners off envy initial. So we conducted with the force. Would be, I think, what into Delta V over there upon the dirty. So much off the ball that is given, that is 2.5 g and the final that is 7.5 given and initially Well, basically the direction of the initial and final velocities approaching this one. So we&#x27;re using right it as a positive. So on the left will be negative. That once manners off 8.5. Obviously. So this makes positive over there. And this will cut down on Delta T with the 12 fives. Again support. Get on the ball. This basically comes out 1 69 1 16 Newton in the right direction

Sean Dougherty · Answer

200 bullets collide with a wall in 10 seconds. He&#x27;s bull is a massive five, no times tending and three kilograms had collided a speed of 1200 meters per second. Now we&#x27;re asked to find the average momentum, the average change momentum per second. So let&#x27;s find that average momentum or the change momentum. I should say the average change momentum tell to pee is equal to the number of bullets times and the final minus him fi initial. And we know that ember the final velocities you records that collide into a wall. So therefore, average change in momentum is just 200 bullets times negative. The mass which is 5.0 times 10 to the negative, three times the initial velocity, 1200 meters per second. And we also want to divide this by, um, 10 seconds because we were asked to find the average momentum change per second. So therefore are change momentum per second is equal to negative 120 kilograms meters per second and anuses per second. Right, so per second squared Essentially, um, now for part B. If we want to find the average force exerted on the wall, we know that the average force, it&#x27;ll just rightist f average sequel to the change of momentum who are until to change of time. Um, and this is essentially exactly what we calculated in part A. Right, Well, we got change momentum, and we know that meant some changes by 120 kilograms per second. And we know our delta t his one based on this because, Mary, we already have momentum per second. And so our momentum is 1 20 per second. So our, um, average force exerted on the wall is just 120 Newtons, no heart. See, we need to find the average pressure, the bullet, because there on the wall. And so this average pressure p average is equal to the average force right over the area exerted on the wall. So it was just plug in these numbers. Some average 40 capital from before 1 20 Newtons Theo area on the wall is given and that&#x27;s just 3.0 times 10 to the negative. Four meters squared. And so our average pressure exerted on the wall is equal to 4.0 times 10 to the five. Paschal&#x27;s

Farhanul Hasan · Answer

are a part. A first. Um, we assume that ball travels in a straight line soon ball travels and straight line. Uh, we also we also it it&#x27;s also a collision where momentum is conserved. Obviously, Um, and we of course, we assume velocity is in the ex direction. Uh, and so, uh, we know that the rate of change of momentum Delta Paedo teasing called the force so f um uh, times Delta TV is the rate of change of momentum and that, um and so and so force is of times old Steve is the change in momentum? Excuse me. So, force is the change in momentum, which is m ah. Visa backs minus minus of loose effects. So it&#x27;s 2 a.m. ah V&#x27;s of ex hoops over a delta T right. The time taken between ah, collisions. So it will be two times uh, 0.58 kilograms. Ah, times are Whoops. I I meant to say I learned to say, um Quinto five a, um 25 8 kilograms. That&#x27;s right. Okay, um times 25 meters per second with speed. And divide that by 0.6 seconds. Uh, and so force is 4.83 Mutants. Importante. We have that. Okay, we first we want to find them distance between the wall and the player. Kind up linked l and so the. And so assuming obviously that there&#x27;s constant speed motion off the ball. You know, the time is it called a distance travel, which is two times the length between the law and that, sir. Aah! Over velocity in the ex directions. So l will be too. Will be books. L will be the X. Pardon me will be v x times t over to, um And so Auntie is delta t from parte eso this 0.6 times 25.6 seconds turns 25 meters per second over too. Gives you, um 7.5 meters. Um and so because he have length and area, we can find out a volume V get us area times length. So that&#x27;s 30 meter square times, 7.5 meters, which he calls, which equals 225 cubic meters. Ah. And so we have this relation from the book that P pressure is 2/3 off n times 1/2 and V squared over volume. Ah, and then and being the number. And so we&#x27;re given that end is 10 ah, 1/2 times 0.58 kilograms. Uh, Times 25 squared over. Ah, the volume which we found to do 2 25 his meter cubed. And so this gives you 0.54 past house.

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Questions 20–22 refer to the following scenario: …

Problem 21 Easy Difficulty

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