Problem 8.23

8.23.

First of all, we need the speed of m₁ just before the collision. We can find this through conservation of energy.

$\frac{1}{2} m_{1} v_{1}^{2} ~=~ m_{1} g h$

$v_{1} = \sqrt{(2)(9.8)(5)} ~=~ 9.90 m/s$

We know from the book that the speed of m₁ just after a perfectly elastic collision is given by

$v_{1f} ~=~ \frac{m_{1}-m_{2}}{m_{1}+m_{2}} v_{1}$

so $v_{1} ~=~ -\frac{1}{3} (9.90 m/s) ~=~ -3.30 m/s$

Now going back to conservation of energy after the collision, to find the highest point that the block rises,

$m_{1} gh_{max} ~=~ \frac{1}{2} m_{1} (-3.30 m/s)^{2}$

or h_max  =  .556 m