A block of mass 1.9\; kg is at rest at the edge of a table, of height 1\; m. A bullet of mass 0.1\; kg collides with the block and sticks to it. If the velocity of the bullet is 20\; m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [ Take g=10\; m/s^{2}. Assume there is no rotational motion and loss of energy after the collision is negligable.]
Option: 1 20\; J
Option: 2 21\: J
Option: 3 19\: J
Option: 4 23\: J

Answers (1)

Let the velocity of the combined system after a collision at A is V

\begin{aligned} &\text { Applying law of conservation of linear momentum }\\ &\Rightarrow 0.1 \times 20=(1.9+0.1) \mathrm{V}\\ &\Rightarrow 2=2 \mathrm{~V}\\ &\Rightarrow V=1 \mathrm{~m} / \mathrm{sec} \end{aligned}

So

At A

\begin{array}{l} \mathrm{KE}=( \frac{1}{2})* \mathrm{mv}^{2}=( \frac{1}{2})* 2*(1)^{2}=1 \mathrm{~J} \\ PE=mgh=2 \times 10 \times 1=20 \ J\\ \mathrm{TE}=\mathrm{KE}+\mathrm{mgh}=1+20=21 \mathrm{~J} \end{array}

As Total energy is conserved 

So

 \\(\mathrm{TE})_A=(\mathrm{TE})_B\\ (\mathrm{TE})_B=21 \ J

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