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Q48PE

Expert-verifiedFound in: Page 291

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A **\(5.50\;kg\) **bowling ball moving at **\(9.00\;{\rm{m/s}}\)**collides with a **\(0.850\;{\rm{kg}}\)**bowling pin, which is scattered at an angle of **\(85.{0^ \circ }\)**to the initial direction of the bowling ball and with a speed of**\(15.0\;{\rm{m/s}}\)**. (a) Calculate the final velocity (magnitude and direction) of the owling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision.**

- The final velocity of the ball is\(8.83\;{\rm{m/s}}\) .
- The collision is inelastic.
- The spin of the ball results the decrease in linear kinetic energy.

The mass of the bowling ball is \({m_1} = 5.50\;kg\)

The mass of the bowling pin is\({m_2} = 0.850\;{\rm{kg}}\).

The speed of the ball is\(v = 9.00\;{\rm{m/s}}\).

The angle of scattering with the initial is\(\theta = 85.0^\circ \).

(a) Using the conservation of momentum along horizontal direction we get,

\(\begin{aligned}{m_1}{v_1}\cos {\theta _1} + {m_2}{v_2}\cos {\theta _2} &= mv\\{v_1} &= \dfrac{{mv - {m_2}{v_2}\cos {\theta _2}}}{{{m_1}\cos {\theta _1}}}\end{aligned}\)

Substituting the values we get,

\(\begin{aligned}{v_1} &= \dfrac{{5.50 \times 9.00 - 0.850 \times 15.0 \times \cos 85.0^\circ }}{{5.50 \times \cos 5^\circ }}\\ &= 8.83\;{\rm{m/s}}\end{aligned}\)

(b) The initial kinetic energy is,

\(\begin{aligned}{\left( {KE} \right)_i} &= \dfrac{1}{2} \times 5.50 \times {9^2}\\ &= 223\;{\rm{J}}\end{aligned}\)

The final kinetic energy is,

\(\begin{aligned}{\left( {KE} \right)_f} &= \dfrac{1}{2} \times 5.50 \times {8.83^2} + \dfrac{1}{2} \times 0.850 \times {15.0^2}\\ &= 310\;{\rm{J}}\end{aligned}\)

As the initial and the final kinetic energies are not equal, the collision is inelastic.

(c) The initial kinetic energy will be more, but as soon as the ball spins a little extra, it allows the ball to gain more speed. For this reason, there will be a decrease in rotational kinetic energy, which results in a decrease in linear kinetic energy.

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