Question

7. Two 1.80 kg bodies, A and B, collide. The velocities before the collision are and . Immediately after the collision block A has . a. Wat is the velocity of block B immediately after the collision? b. What is the change in the total kinetic energy because of the collision? ((please make the solution readable ))

Answer 1

The speeds are missing, but I'll answer assuming random values (for initial speeds and A speed after the collision):

The conservation of momentum is given by:

$\frac{(1.80kg)(2m/s + 4m/s)}{(1.80kg)} - 1.5m/s = v'_{2}$

ANSWER

b.-

The change in kinetic energy is:

$\Delta KE = \frac{1}{2}m_{1}(v'_{1})^{2} + \frac{1}{2}m_{1}(v'_{2})^{2} - \frac{1}{2}m_{1}(v_{1})^{2} - \frac{1}{2}m_{2}(v_{2})^{2}$

$\Delta KE = \frac{1}{2}(1.80kg)\left ((1.5m/s)^{2} + (4.5m/s)^{2} - (2m/s)^{2} - (4m/s)^{2} \right )$

ANSWER $\Delta KE = 2.25J$

I hope it helps!!