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Question

Physics Question on work, energy and power

A block CC of mass m is moving with velocity vo{{v}_{o}} and collides elastically with block AA of mass m and connected to another block BB of mass 2m2m through spring of spring constant kk. What is kk, if xox_o is compression of spring when velocity of AA and BB is same?

A

mv02x02\frac{m v_{0}^{2}}{x_{0}^{2}}

B

mv022x02\frac{m v_{0}^{2}}{2x_{0}^{2}}

C

32mv02x02 \frac{3}{2} \frac{m v_{0}^{2}}{x_{0}^{2}}

D

23mv02x02 \frac{2}{3} \frac{m v_{0}^{2}}{x_{0}^{2}}

Answer

23mv02x02 \frac{2}{3} \frac{m v_{0}^{2}}{x_{0}^{2}}

Explanation

Solution

Using conservation of linear momentum, we have
mv0=mv+2mvmv_0 = mv + 2 mv
v=v03\Rightarrow v = \frac{ v_0 }{ 3}
Using conservation of energy, we have
12mv02=12kx02+12(3m)v2\frac{1}{2} mv_0^2 = \frac{1}{2} kx_0^2 + \frac{1}{2} ( 3m)v^2
where x0x_0 is compression in the string.
mv02=kx02+(3m)v02g\therefore mv_0^2 = kx_0^2 + ( 3 m ) \frac{ v_0^2 }{ g }
kx02=mv02mv023\Rightarrow kx_0^2 = mv_0^2 - \frac{ mv_0^2 }{ 3}
kx02=2mv023\Rightarrow kx_0^2 = \frac{ 2 mv_0^2 }{ 3 }
k=2mv023x02\therefore k = \frac{ 2 mv_0^2 }{ 3 x_0^2 }