00:01
In this problem we have given starting with an initial speed 5 meter per second we have given a diagram like this this is a ball of mass 4 .60 kg and here another ball of mass 1 .50 kg this height is given 0 .30 meter and initial speed is given 5 meter per second so in a we have to find the speed of 1 .50 kg ball just before impact.
00:50
So here we will use energy conservation between point a and point b.
00:55
Energy at point a should be equal to energy at point b.
01:00
So kinetic energy will be equal to half mass into speed square plus potential energy m into g which is 9 .81 into h which is 0 .30 should be equal.
01:15
To energy at point b.
01:16
Here potential energy is 0 plus current energy half m v .a square.
01:27
Right? so this va is the speed of ball a before impact.
01:35
So from here we can calculate the value of ba square will be equal to 5 square plus 2 into 9 .81 into 0.
01:49
0 .30.
01:51
So this will be equal to under root 25 plus 2 into 9 .81 into 0 .300.
02:07
So this is equal to 5 .57 meter per second.
02:15
This is the speed of ball having mass 1 .50 k .g before the impact.
02:22
Now in part b we have to find the velocities.
02:27
Of both balls just after the collision now this is ball of 1 .50k it has speed before impact 5 .57 and here another ball which has mass 4 .60kg vb suppose this is ball b this is ball b this is ball a va initial is va initial is va initial is 5 .57 meter per second and vb initial is 0.
03:15
So here we will write momentum conservation equation.
03:22
Momentum will be concerned.
03:24
P initial will be equal to p final.
03:26
Suppose va final is the speed of ball a after the collision and vb final is the speed...