Inelastic collisions in one dimension: A 5.00-kg ball is hanging from a long but very light flexible wire when it is struck by a 1.50-kg stone traveling horizontally to the right at 12.0 m/s. The stone rebounds to the left with a speed of 8.50 m/s, and the ball swings to a maximum height h above its original level. The value of h is closest to

Question

Quizplus · Accepted Answer

First, we need to find the velocity of the ball and stone immediately after collision. By conservation of momentum in one dimension:
$m_1v_{1i}+m_2v_{2i}=(m_1+m_2)v_f$
where $m_1$ and $v_{1i}$ are the mass and initial velocity of the ball, $m_2$ and $v_{2i}$ are the mass and initial velocity of the stone, and $v_f$ is the final velocity of both objects after collision.
Plugging in the known values, we get:
$(5.00 \ kg)(0 \ m/s)+(1.50 \ kg)(12.0 \ m/s)=(5.00 \ kg+1.50 \ kg)v_f$
Solving for $v_f$, we get $v_f=2.67 \ m/s$.
Next, we can use conservation of energy to find the height the ball reaches. The total initial energy of the system (just before the collision) is:
$E_i=\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2$
The final energy of the system (at the maximum height of the ball) is:
$E_f=\frac{1}{2}(m_1+m_2)v_f^2+m_1gh$
where $g$ is the acceleration due to gravity, and $h$ is the maximum height reached by the ball.
By conservation of energy, $E_i=E_f$, so we can set the two equations equal to each other and solve for $h$:
$\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}(m_1+m_2)v_f^2+m_1gh$
Plugging in the known values, we get:
$\frac{1}{2}(5.00 \ kg)(0 \ m/s)^2 + \frac{1}{2}(1.50 \ kg)(12.0 \ m/s)^2 = \frac{1}{2}(5.00 \ kg + 1.50 \ kg)(2.67 \ m/s)^2 + (5.00 \ kg)(9.81 \ m/s^2)h$
Solving for $h$, we get $h=1.93 \ m$, which is closest to choice C.

Inelastic Collisions in One Dimension: a 5