P01G2007 – Balls Colliding in Mid-air
Balls Colliding Mid-air
A ball is thrown straight up from the ground with initial speed $v_0$. At the same instant, a second ball is dropped from rest from a height $H$, directly above the point where the first ball was thrown upward. In terms of the given variables and $g$, the acceleration due to gravity, find.
1. The time at which the two balls collide.
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The first ball is launched upward with velocity $v_0$ from an initial height $y_0=0$ and the kinematic equations describing its motion are therefore
\left\{ \begin{array}{c}
\displaystyle{y_1\left(t\right)=-\frac{1}{2}gt^2+v_0t} \\
\\
v_1\left(t\right)=-gt+v_0 \ \ \ \ \ \end{array}
\right.The second ball is dropped from rest $\left(v_0=0\right)$ from an initial height $y_0=H$ and the kinematic equations describing its motion are therefore
\left\{ \begin{array}{c}
\displaystyle{y_2\left(t\right)=-\frac{1}{2}gt^2+H} \\
\\
v_2\left(t\right)=-gt \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}
\right.The balls collide at time $t_c$ such that $y_1\left(t_c\right)=y_2\left(t_c\right)$ and we solve for $t_c$ as follows
\begin{aligned}
y_1\left(t_c\right)=y_2\left(t_c\right)\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_c+v_0t_c=-\frac{1}{2}gt^2_c+H \\
\\
&\Rightarrow \ \ \ \ \ t_c=\frac{H}{v_0}
\end{aligned}Thus, the time $t_c$ at which the balls collide is equal to
\boxed{t_c=\frac{H}{v_0}}2. The height above the ground where they collide.
View answerHide answerWhen the balls collide, they have the same height which is equal to
\boxed{y_1\left(t_c\right)=y_2\left(t_c\right)=-\frac{1}{2}gt^2_c+H=-\frac{gH^2}{2v^2_0}+H}3. The velocity of each ball just before the collision.
View answerHide answerThe velocity of each ball just before the collision is equal to
\boxed{
\begin{aligned}
v_1\left(t_c\right)&=-gt_c+v_0=-\frac{gH}{v_0}+v_0 \\
v_2\left(t_c\right)&=-gt_c=-\frac{gH}{v_0}
\end{aligned}
}4. The relative speed between the two balls just before the collision.
View answerHide answerThe relative speed between the two balls just before the collision is equal to
\boxed{v_{rel}=\left|v_1\left(t_c\right)-v_2\left(t_c\right)\right|=\left|-\frac{gH}{v_0}+v_0-\left(-\frac{gH}{v_0}\right)\right|=v_0}