Answer
(a) The speed of the 2.50-kg block is 3.98 m/s
The speed of the 7.00-kg block is 4.42 m/s
(b) The block slides a distance of 1.62 meters up the incline.
Work Step by Step
(a) Note that $m = 2.50~kg$ and $M = 7.00~kg$.
We need to find $v_A$, the speed of block $m$ at the bottom of the incline. We can use conservation of energy to find the speed:
$KE = PE$
$\frac{1}{2}mv_A^2 = mgh$
$v_A = \sqrt{2gh} = \sqrt{(2)(9.80~m/s^2)(3.60~m)}$
$v_A = 8.4~m/s$
We can use conservation of momentum to set up an equation:
$m~v_A + 0 = m~v_A' + M~v_B'$
Since the collision is elastic, we can use Equation 7-7 to set up another equation:
$v_A - 0 = v_B' - v_A'$
$v_A' = v_B' - v_A$
We can use this expression for $v_A'$ in the first equation:
$m~v_A + 0 = m~v_B' - m~v_A + M~v_B'$
$v_B' = \frac{2m~v_A}{m+M} = \frac{(2)(2.50~kg)(8.4~m/s)}{(2.50~kg)+(7.00~kg)}$
$v_B' = 4.42~m/s$
We can use $v_B'$ to find $v_A'$:
$v_A' = v_B' - v_A$
$v_A' = 4.42~m/s - 8.4~m/s = -3.98~m/s$
Note that the negative sign means that the small block is going back up the slope.
(b) We can use conservation of energy to find the final height of the small block.
$PE = KE$
$mgh = \frac{1}{2}m(v_A')^2$
$h = \frac{(v_A')^2}{2g} = \frac{(-3.98~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 0.808~m$
We can use the height $h$ to find the distance $d$ that the small block slides up the incline.
$sin(\theta) = \frac{h}{d}$
$d = \frac{h}{sin(\theta)} = \frac{0.808~m}{sin(30^{\circ})}$
$d = 1.62~m$
The block slides a distance of 1.62 meters up the incline.
