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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A ball of mass m strikes an identical ball at rest with speed of 10 m/s. After collision which is assumed perfectly elastic the first ball is found to move with speed 8 m/s. Speed of $2 ^{nd}$ ball and angle between the paths of two balls after collision is

\[(a)\;4\; m/s; 45 ^{\circ} \quad (b)\;6\; m/s; 45 ^{\circ} \quad (c)\;4\; m/s; 90 ^{\circ} \quad(d)\;6\; m/s; 90 ^{\circ} \]

1 Answer

We use conservation of kinetic energy as we cannot use conservation of linear momentum as direction of the ball after collision is not known
KE before collision $=\large \frac{1}{2} $$m .10 ^2+0$
$\qquad=50 \;mJ$
KE after collision $=\large \frac{1}{2} $$m \times 8 ^2+\large\frac{1}{2}$$ m \; v^2$
$\qquad=(32 +\large\frac{1}{2}$$mv^2)J$
$32\;m+\large\frac{1}{2}$$mv^2=50 m$
$v=6$$ m/s$
Since we see that $10^2=8^2+6^2$
The two balls must be moving at right angles to each other.
Hence d is the correct answer
answered Jul 5, 2013 by meena.p
edited May 26, 2014 by lmohan717

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