Question

a block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equal parts continuing in the same direction. if one of part moves at the speed of 30 m/s, with what speed does the second part move and what is the frictional change in the kinetic energy​

Answer 1

Given:

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• A block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equal parts continuing in the same direction.

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To find:

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The frictional change in the kinetic energy if one of the parts moves at 30 m/s with what a speed does the second part move.

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Solution:

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Here it is given that the block is moving horizontally on the smooth surface and there is no external force on the block. There is no external force acting. The internal force breaks the block in two parts. We know that, p = mv.

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Here linear momentum of the block, before the break should be equal to the linear momentum after two parts after the break.

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And all according to the question,

all velocities are in the same direction. So,

M(20 m/s) = m/2 (30 m/s) + m/2 × v( speed of the other part)

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==> M(20 m/s) = m/2 (30 m/s) + m/2 × v

==> v = 10 m/s

Now, the change in kinetic energy will be

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1/2 × M/2 (30 m/s)$^{2}$ + 1/2 × M/2 (10 m/s)$^{2}$ – 1/2 M(20 m/s)$^{2}$

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= M/2 (450 + 50 – 400) m$^{2}$ / s$^{2}$

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$= ( 50 \frac{m {}^{2} }{s {}^{2} } )M$

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We were asked to find fractional change in the kinetic energy.

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So, fractional change in the kinetic energy will be

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$\frac{M ( 50\frac{m {}^{2} }{s {}^{2} } )}{ \frac{1}{2} M (20 \: \frac{m}{s } ) {}^{2} }$

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Thus, we will get 1/4.

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Therefore, fractional change in the kinetic energy is 1/4.