Question

Q4. A 5.0-kg bowling ball rolls down africtionless 2.5 m tall ramp and strikes astationary mass at the bottom of the ramp in aperfectly elastic collision. To whatheight back up the ramp does the first bowlingtravel after the collision ita) the stationary mass is also 5.0 kgb) the stationary mass is 10.0 kgC) the stationary mass is 500.00 kg

Answer 1

a)

Using conservation of energy for ball 1:

$\begin{gathered} E1=E2 \\ K1+U1=K2+U2 \\ 0+m1gh=\frac{1}{2}m1v1^2 \end{gathered}$

Solve for v1:

$\begin{gathered} v1=\sqrt[]{2gh} \\ v1=\sqrt[]{2(9.8)(2.5)} \\ v1=7\frac{m}{s} \end{gathered}$$v1=v2^{\prime}-v1^{\prime}$

Using conservation of momentum:

$\begin{gathered} m1v1=m1(v2^{\prime}-v1^{\prime})+m2v2^{\prime} \\ v2^{\prime}=\frac{2m1v1}{(m1+m2)} \\ \end{gathered}$

a)

m2 = 5 kg

$v2^{\prime}=7\frac{m}{s}$

so:

$\begin{gathered} v1^{\prime}=7-7 \\ v1^{\prime}=0 \end{gathered}$

so:

$\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}=0 \end{gathered}$

If the stationary mass is 5.0 kg the height back up the ramp is 0 meters

b)

m2 = 10

$v2^{\prime}=\frac{14}{3}\frac{m}{s}$

so:

$v1^{\prime}=-\frac{7}{3}\frac{m}{s}$$\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx0.2778m \end{gathered}$

If the stationary mass is 10.0 kg the height back up the ramp is 0.2778 meters

c)

m2 = 500.00kg

$v2^{\prime}=\frac{14}{101}\frac{m}{s}$

so:

$v1^{\prime}\approx-\frac{693}{101}\frac{m}{s}$$\begin{gathered} m1gh^{\prime}=\frac{1}{2}m1(v1^{\prime})^2 \\ h^{\prime}\approx2.4m \end{gathered}$

If the stationary mass is 500.0 kg the height back up the ramp is 2.4 meters