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Standard XII

Physics

Oblique Collision

Two masses ...

Question

Two masses $m_{1} = 2 k g$ and $m_{2} = 5 k g$ are moving on a directionless surface with velocities 10 m/s and 3 m/s respectively $m_{2}$ is ahead of $m_{1}$ . An ideal spring of spring constant $k = 1120 N / m$ is attached on the back side of $m_{2}$ . The maximum compression of the spring will be, if on collision the two bodies stick together is approximately:

0.51 m

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0.062 m

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0.3 m

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0.72 m

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Solution

The correct option is C $0.3 m$
When they stick together momentum will be conserved.

$2 \times 10 + 3 \times 5 = (2 + 5) \times v$ i.e. $v = 5 m / s$

Apply law of conservation of energy.

$\frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2} = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} k x^{2}$

$k x^{2} = 2 \times 10 \times 10 + 3 \times 5 \times 5 - 7 \times 5 \times 5 = 100$

$x^{2} = \frac{100}{1120}$

$x = \frac{1}{\sqrt{11.2}} \approx 0.3 m$

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The correct option is C 0.3mWhen they stick together momentum will be conserved.2×10+3×5=(2+5)×v i.e. v=5m/sApply law of conservation of energy.12m1v12+12m2v22=12(m1+m2)v2+12kx2kx2=2×10×10+3×5×5−7×5×5=100x2=1001120x=1√11.2≈0.3m