Need all 4 1 a 4 kg cart moving to the east at 5 ms collides...

Need all 4 1. A 4 kg cart moving to the east at 5 m/s collides inelastically with a 3 kg cart moving to the west at 4 m/s. The carts are on an air track, so friction is not a factor. What is the final velocity of the carts? 2. A 1.2 kg skateboard is coasting down a sidewalk at 5 m/s when a .8 kg cat drops vertically from a tree on to the skateboard. What is the new velocity of the board with the cat on it? 3. A 7500 kg train car travelling 2 m/s to the right hits and hooks to a 9000 kg train car moving .75 m/s to the left. What is the final velocity of the cars? How much KE was lost in the collision? 4. A .025 kg marble moving to the right at .2 m/s collides elastically with a .01 kg marble also moving to the right at .15 m/s. If the final speed of the second marble is .22 m/s, what is the final speed of the first marble?

Transcript

00:01 Question number one, a cart of mass m1 equal to 4 kg is moving with a velocity u1 equal to 5 meter per second towards east.

00:22 It hits the another cart of mass m2 equal to 3 kg moving with a velocity u2 4 meter per second towards west we need to find out the velocity their final velocity after the collision after the collision they move like a common body so total linear momentum before collision is the linear momentum of cart of mass 4 kggm 1 u1 and plus cart of mass 3 kgm 2 u2 after the collision they become one piece so there linear momentum will be m1 plus m2 into v.

01:19 Right? we put the value m1 is equal to 4 kg, u1 5 meter per second, m2 equal to 3 kg, and u2 is equal to minus 4 meter per second because u1 is towards east, we are taking it positive.

01:37 So towards west, we should take it as a negative.

01:40 So this equal to m1, we put here value of m1 and m2, m1 is 4 plus 3 into v.

01:55 If we solve this, we get v equal to 1 .14 meter per second.

02:03 The positive sign indicate it is same direction like u1 means it is towards east.

02:10 So after the collision, they move towards east.

02:16 In part two.

02:19 This is a skat car.

02:25 This is a skate car and a skate car is moving like this.

02:34 And the cat falls on it.

02:38 This is the cat which falls vertically down.

02:45 So in this problem, we need to apply the law of conservation of linear momentum in the horizontal direction because cart is moving in the horizontal direction.

02:58 This is cat and this is cat board.

03:12 So we apply the total linear momentum of the cart cat in the before the cat falls on the skateboard.

03:22 This is the one incidence and second is when the cat has already hit the skateboard.

03:30 So before means before the cat hits the skateboard and after means after the cat hits the skateboard.

03:37 Board and both of them move together.

03:39 So before and after, we apply the law of conservation of linear momentum in means horizontal direction.

03:50 This is also means horizontal direction, right? so before now the initial linear momentum of the cat is completely means vertical.

04:04 So it's means horizontal component is zero.

04:07 So before the cat falls on the skateboard, it's.

04:11 Means horizontal linear momentum is zero and for skateboard it is moving in the horizontal direction so its linear momentum will be its mass 1 .2 kg and plus multiplied by the velocity with it is moving and it should be equal to after now after both cat and skateboard are moving in the horizontal direction with a combined velocity v so their linear moment will be equal to m1 plus m2 multiplied by v.

04:46 It means mass of the skateboard 1 .2 kg, mass of the cat 0 .8 kg multiplied by v.

04:54 So we solve it, the velocity is 3 meter per second...

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