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2 years ago

7

A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a

distance of 0.38 cm. Find the speed of the bullet as it emerges from the block if its initial speed is 450 m/s

1 answer:

o-na [289]2 years ago

4 0

Answer:395.6 m/s

Explanation:

Given

mass of bullet $m=5 gm$

mass of wood block $M=1 kg$

Length of string $L=2 m$

Center of mass rises to an height of $0.38 cm$

initial velocity of bullet $u=450 m/s$

let $v_1$ and $v_2$ be the velocity of bullet and block after collision

Conserving momentum

$mu=mv_1+Mv_2$ -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

$\frac{Mv_2^2}{2}=Mgh_{cm}$

$v_2=\sqrt{2gh_{cm}}$

$v_2=\sqrt{2\times 9.8\times 0.38}$

$v_2=0.272 m/s$

substitute the value of $v_2$ in equation 1

$5\times 450=5\times v_1+1000\times 0.272$

$v_1=395.6 m/s$

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