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Standard XII

Physics

Relation between Linear and Angular Speed

A 250 g grass...

Question

A 250 g grasshopper moving due south at 20 cm $^{- 1}$ (at mid air) collides with 150 g grasshopper moving 60cms $^{- 1}$ due north. Find the decrease in KE if they move together after collision.

0.3 J

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3.0 J

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0.03 J

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0.003 J

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Solution

The correct option is C 0.03 J
$△ K E = \frac{m_{1} m_{2}}{2 (m_{1} m_{2})} (v_{1} - v_{2})^{2}$
$= \frac{0.25 \times 0.15 (0.8)^{2}}{2 (0.4)} = 0.03 J$

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A 250 g grasshopper moving due south at 20 cm−1 (at mid air) collides with 150 g grasshopper moving 60cms−1 due north. Find the decrease in KE if they move together after collision.

The correct option is C 0.03 J△KE=m1m22(m1m2)(v1−v2)2=0.25×0.15(0.8)22(0.4)=0.03J

A 250 g grasshopper moving due south at 20 cm $^{- 1}$ (at mid air) collides with 150 g grasshopper moving 60cms $^{- 1}$ due north. Find the decrease in KE if they move together after collision.

The correct option is C 0.03 J
$△ K E = \frac{m_{1} m_{2}}{2 (m_{1} m_{2})} (v_{1} - v_{2})^{2}$
$= \frac{0.25 \times 0.15 (0.8)^{2}}{2 (0.4)} = 0.03 J$