Consider the accompanying diagram. Ball A is allowed to fall...

Consider the accompanying diagram. Ball A is allowed to fall and strike ball B. Assume that all of ball A's energy is transferred to ball $\mathrm{B}$ at point I, and that there is no loss of energy to other sources. What is the kinetic energy and the potential energy of ball $\mathrm{B}$ at point II? The potential energy is given by $\mathrm{PE}=m g z,$ where $m$ is the mass in kilograms, $g$ is the gravitational constant $\left(9.81 \mathrm{m} / \mathrm{s}^{2}\right),$ and $z$ is the distance in meters.

Transcript

00:01 So in this question we need to calculate the potential energy.

00:04 There are two balls a and b.

00:08 What they said that the a strikes on b, the ball a strikes on b, and we have to calculate the difference in potential energy.

00:16 So we'll do it.

00:19 Number one, the potential energy of ball a at starting or starting position.

00:31 So we know formula potential energy is equal to mass into 1.

00:34 Gravity acceleration due to gravity into height mgh so this found to be 2 kg into 9 .8 meter per second square into 10 height is 10 meter so this found to be 196 jule this is the potential energy of ball a we'll write it as a potential energy of a this is the first then in second part we have to calculate the potential energy of b of b of b of b b at starting as well as after some time.

01:12 At starting.

01:14 So as we know what we said that a strikes on b, it means the b will be at rest.

01:20 Therefore the potential energy, starting potential energy, we can write pe starting or potential energy as starting is zero.

01:32 As before striking it was at rest.

01:35 We can write as it was rest at rest.

01:42 Or it was a test.

01:44 Then the potential energy of b, this point b, at second position, at second position.

01:54 This also we can calculate with the same method, that is the potential energy of b at 2.

02:01 Again the formula m g h...

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