00:01
Hello students we are given a question and in this question there is two there are two cars so first car is represented by m1 so its mass is 1 .13 multiplied by 10 to the power 3 kg its velocity u1 is 25 .7 meter per second and it is towards the east mass of the second car is 1 .25 multiplied by 10 to the power 3 kg and and its initial speed is minus 13 .8 meter per second because it is towards the west.
00:38
So we consider the speed or the velocity towards the west as negative.
00:44
Now in the a part we have to find a common speed after the collision and it is given that they get attached after the collision.
00:53
So we will basically use here conservation of momentum.
01:01
Term so we will get m1 u1 plus m2 u 2 is equal to m1 plus m2 v so it will be equal to 1 .13 multiplied by 10 to the power 3 multiplied by 25 .7 plus m 2 which is 1 .25 multiplied by u 2 which is minus 13 .8 is equal to m1 plus m2 so it will be 1 .13 multiplied by 10 to the power 3 plus 1 .25 multiplied by 10 to the power 3 multiplied by v.
01:40
So here the velocity v comes out to be 4 .954 meter per second.
01:46
So as it is positive it means the whole system move towards the east.
01:52
So this is the velocity after the collision.
01:55
Now in the b part we have to tell what will be the change in the momentum of each car.
02:01
So change in the momentum of the 4.
02:03
First car will be m1 v1 minus m1 u1 so it will be 1 .13 multiplied with 10 to the bar 3 multiplied with 4 .9 5 4 minus 25 .7 so the change in momentum is minus 2 3 3 422 .9 kilogram meter per second and for the second car p2 is equal to m2 multiplied with v here it will be v or in this case it will also be v because final speed is always v because they move together so it will be v minus v2 so it will be 1 .2725 multiplied with 10 to the power 3.
02:57
V is equal to 4 .954 minus of minus 13 .8 so it comes out to equal to 23...