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Q43PE

Expert-verifiedFound in: Page 291

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of **\(10.0kg\)**and the horizontal component of its velocity is **\(8.00 m/s\)**when the **\(65.0kg\)**performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity?**

The velocity of the system after collision will be \(1.07 m/s\).

**According to collision theory, only a specific number of collisions between acceptable reactant particles with the correct orientation result in a detectable or noticeable change; these successful modifications are referred to as successful collisions.**

The velocity of cannon ball is \({v_c} = 8.00 m/s\)in right side.

The mass of the cannon ball is\({M_c} = 10.0kg\).

The velocity of the person is\({v_p} = 0\)

The mass of the person is\({M_p} = 65kg\)

The Collison is inelastic.

By putting all the value into the equation we get

\(\begin{aligned}{M_c}{V_c} + {M_p}{V_p} &= + {M_T}{V_f}\\{V_f} &= \dfrac{{{M_v}{V_c} + {M_p}{V_p}}}{{{M_T}}}\\{V_f} &= \dfrac{{\left( {10} \right)\left( 8 \right) + \left( {65} \right)\left( 0 \right)}}{{\left( {10 + 65} \right)}}\\{V_f} &= 1.07\,m/s\end{aligned}\)…………………(1)

Hence the velocity of the system after collision will be\(1.07 m/s\). They will move towards right side as the velocity answer is positive.

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