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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

User David Lichteblau
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1 Answer

21 votes
21 votes

a)

Using conservation of energy for ball 1:


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

Solve for v1:


\begin{gathered} v1=\sqrt[]{2gh} \\ v1=\sqrt[]{2(9.8)(2.5)} \\ v1=7(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)=(2m1v1)/((m1+m2)) \\ \end{gathered}

a)

m2 = 5 kg


v2^(\prime)=7(m)/(s)

so:


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

so:


\begin{gathered} m1gh^(\prime)=(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)=(14)/(3)(m)/(s)

so:


v1^(\prime)=-(7)/(3)(m)/(s)
\begin{gathered} m1gh^(\prime)=(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)=(14)/(101)(m)/(s)

so:


v1^(\prime)\approx-(693)/(101)(m)/(s)
\begin{gathered} m1gh^(\prime)=(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

User Volodymyr Usarskyy
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2.9k points