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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

A block of mass $0.50\;kg$ is moving with speed of $2.00 m/s$ on a smooth surface. It strikes another mass of $1.00 kg$ at rest and they move together as a single body. The energy loss during collision is

\[(a)\;0.67 \;J\quad (b)\;1.00 \;J \quad (c)\; 0.16 \;J \quad (d)\;0.34 \;J\]
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1 Answer

$m$ -mass of $1st$ body, $M$ -Mass of the Initial kinetic energy of system
$K_i=\large\frac{1}{2}$$mu^2+\large\frac{1}{2}$$M(0)^2$
$\qquad=\large\frac{1}{2} $$ \times 0.5 \times 2^2$
$\qquad=1 J$
For collision applying conservation of linear momentum
$mu=(M+m)v$
$0.5 \times 2=(0.5 +1)v$
$v=\large\frac{2}{3}$$ m/s$
Finally Kinetic energy
$K_f=\large\frac{1}{2}$$(0.5+1) \times \large\frac{2}{3} \times \frac{2}{3}$
$\qquad=\large\frac{1}{3}\;$$J$
Energy lost $=1-\large\frac{1}{3}=\frac{2}{3}$$=0.67 \;J$
Hence a is the correct answer.
 

 

answered Aug 7, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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