00:01
Hi, here in this given problem there is a frictionless incline plane, a ramp whose topmost point is at a height, initial height of edge.
00:22
And this height, h, is given as 3 .1 meter.
00:28
And then its base that is a rough surface a rough horizontal surface.
00:40
A block has been kept over here at the ground and a block starts coming from this top.
00:51
This mass m1, this mass m2.
00:56
And this m2 is related with m1 as four times.
01:01
Of m1 and this the second block initially that was at rest so we can say v2i is 0 and this v1 i means initially speed of the first block when it will come to the ground and it will hit the second block it will be having a speed using conservation of energy we can say this v1 i when it will come to the ground.
01:31
Its initial speed will be given by square root of 2gh.
01:35
And plugging in the known values, this is 2 times of 9 .8 into 3 .1.
01:42
And this is calculated to be equal to 7 .8 meter per second.
01:51
After which this second block moves to a distance d and comes to rest.
02:01
Means its final speed that is to be zero and we have to find this distance traveled by the second block before coming to rest provided the two conditions in the first condition the collision collision of m1 and m2 if it is elastic and in case of elastic collision we know the expression for the final speed gained by the second block v2f, that is given as m2 minus m1 into its initial speed v2i plus twice of m1 into v1i, which we have just found and divided by m1 plus m2.
03:01
But as v2i is 0, so this first term will come out to be equal to 0.
03:07
So this v2f will be given as this will be 0 in the first term plus twice of m1 into 7 .8 and denominator m1 plus m2 means m1 plus 4 m1 means this is 2 times of 7 .8 m1 divided by 5m1 and finally this speed comes out to be equal to 3 .12 meter per second...