JEE Advance - Physics (2019 - Paper 2 Offline - No. 13)
A ball is thrown from ground at an angle $$\theta $$ with horizontal and with an initial speed u0. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is V1. After hitting the ground, the ball rebounds at the same angle $$\theta $$ but with a reduced speed of $${{{u_0}} \over \alpha }$$. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8 V1, the value of $$\alpha $$ is ..................
پاسخ دهید
4.0
توضیح
For first projectile,
$$ < V > = {R \over T} = {U_x} = {v_1}$$
For journey,
$$ < V{ > _{1\,to\,n}} = {{{R_1} + {R_2} + ... + {R_n}} \over {{T_1} + {T_2} + ... + {T_n}}}$$
$$ = {{{{2{u_{{x_1}}} + 2{u_{{y_1}}}} \over g} + {{2{u_{x2}} + 2{u_{{y_2}}}} \over g} + ... + {{2{u_{{x_n}}} + 2{u_{{y_n}}}} \over g}} \over {{{2{u_{{y_1}}}} \over g} + {{2{u_{y2}}} \over g} + ...{{2{u_{{y_n}}}} \over g}}}$$
$${u_x}\left[ {{{1 + {1 \over {{\alpha ^2}}} + {1 \over {{\alpha ^4}}} + ...{1 \over {{\alpha ^{2n}}}}} \over {1 + {1 \over \alpha } + {1 \over {{\alpha ^2}}} + ...{1 \over {{\alpha ^n}}}}}} \right] = 0.8{v_1}$$
$${{{v_0}\left[ {{1 \over {1 - {1 \over {{\alpha ^2}}}}}} \right]} \over {\left[ {{1 \over {1 - {1 \over \alpha }}}} \right]}} = 0.8{v_1}$$
$$ \Rightarrow {\alpha \over {1 + \alpha }} = 0.8$$
$$ \Rightarrow \alpha = 4$$
$$ < V > = {R \over T} = {U_x} = {v_1}$$
For journey,
$$ < V{ > _{1\,to\,n}} = {{{R_1} + {R_2} + ... + {R_n}} \over {{T_1} + {T_2} + ... + {T_n}}}$$
$$ = {{{{2{u_{{x_1}}} + 2{u_{{y_1}}}} \over g} + {{2{u_{x2}} + 2{u_{{y_2}}}} \over g} + ... + {{2{u_{{x_n}}} + 2{u_{{y_n}}}} \over g}} \over {{{2{u_{{y_1}}}} \over g} + {{2{u_{y2}}} \over g} + ...{{2{u_{{y_n}}}} \over g}}}$$
$${u_x}\left[ {{{1 + {1 \over {{\alpha ^2}}} + {1 \over {{\alpha ^4}}} + ...{1 \over {{\alpha ^{2n}}}}} \over {1 + {1 \over \alpha } + {1 \over {{\alpha ^2}}} + ...{1 \over {{\alpha ^n}}}}}} \right] = 0.8{v_1}$$
$${{{v_0}\left[ {{1 \over {1 - {1 \over {{\alpha ^2}}}}}} \right]} \over {\left[ {{1 \over {1 - {1 \over \alpha }}}} \right]}} = 0.8{v_1}$$
$$ \Rightarrow {\alpha \over {1 + \alpha }} = 0.8$$
$$ \Rightarrow \alpha = 4$$