A smooth spherical ball strikes a smooth horizontal floor at an angle <math> <mi>θ</mi> <mo>=</mo> <msup> <mrow> <mrow> <mn>45</mn> </mrow> </mrow> <mrow> <mrow> <mo>?</mo> </mrow> </mrow> </msup> </math> . The coefficient of restitution between the ball and the floor is <math> <mi>e</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)

Vertical component of velocity just after collision = u 2 2 k i = m u 2 2 k f = 1 2 m u 2 2 + 1 2 m u 2 8 = 5 m u 2 16 Fraction = k i - k f k i = 1 - 5 m u 2 16 m u 2 2 = 1 - 5 8 = 3 8

A smooth spherical ball strikes a smooth horizontal floor at an angle <math> <mi>θ</mi> <mo>=</mo> <msup> <mrow> <mrow> <mn>45</mn> </mrow> </mrow> <mrow> <mrow> <mo>?</mo> </mrow> </mrow> </msup> </math> . The coefficient of restitution between the ball and the floor is <math> <mi>e</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision) &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Kinetic Energy Work, Energy and Power Physics Class 11th

A smooth spherical ball strikes a smooth horizontal floor at an angle $θ = 45^{?}$ . The coefficient of restitution between the ball and the floor is $e = \frac{1}{2}$ . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)

Option 1 -

$\frac{5}{8}$

Option 2 -

$\frac{3}{8}$

Option 3 -

$\frac{3}{4}$

Option 4 -

$\frac{1}{4}$

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Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 2

Detailed Solution:

Vertical component of velocity just after collision $= \frac{u}{2 \sqrt[]{2}}$

$k_{i} = \frac{m u^{2}}{2} k_{f} = \frac{1}{2} m \frac{u^{2}}{2} + \frac{1}{2} m \frac{u^{2}}{8} = \frac{5 m u^{2}}{16}$ $\frac{}{}$

Fraction $= \frac{k_{i} - k_{f}}{k_{i}} = 1 - \frac{\frac{5 m u^{2}}{16}}{\frac{m u^{2}}{2}} = 1 - \frac{5}{8} = \frac{3}{8}$

Vertical component of velocity just after collision <math> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mroot> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow></mrow> </mroot> </mrow> </mrow> </mfrac> </math> <math> <msub> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo></mo> <msub> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>m</mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>m</mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>5</mn> <mi>m</mi> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </mfrac> </math> <math><mfrac><mrow><mrow></mrow></mrow><mrow><mrow><mroot><mrow><mrow></mrow></mrow></mroot></mrow></mrow></mfrac></math>Fraction  <math> <mo>=</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>-</mo> <mfrac> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>5</mn> <mi>m</mi> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>u</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </mfrac> </math> <div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1757488991phpegTwA8.jpeg" alt="" width="240" height="116" loading="lazy"></picture></div></div>

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A smooth spherical ball strikes a smooth horizontal floor at an angle $θ = 45^{?}$ . The coefficient of restitution between the ball and the floor is $e = \frac{1}{2}$ . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)

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A smooth spherical ball strikes a smooth horizontal floor at an angle θ = 45 ? . The coefficient of restitution between the ball and the floor is e = 1 2 . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)

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A smooth spherical ball strikes a smooth horizontal floor at an angle $θ = 45^{?}$ . The coefficient of restitution between the ball and the floor is $e = \frac{1}{2}$ . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)