A block of mass 2kg is projected with a velocity 6m/s so that it climbs onto z smooth wedge of mass 1kg. If the block does not leave the wedge, find the maximum height attained by the block. Take g=10m/s2.
A
0.3m
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B
0.6m
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C
1.2m
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D
0.4m
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Solution
The correct option is B0.6m The block attains a maximum height when the final relative velocity between the block and wedge becomes zero. Final velocity of wedge is equal to final velocity of the block.
Kinetic energy of the system with respect to COM is given by Kcom=12m1m2(m1+m2)(v1−v2)2
Final kinetic energy of system with respect to COM is (Kcom)final=12m1m2(m1+m2)(v1f−v2f)2 (Kcom)final=0 [∵v1f=v2f]
and Initial kinetic energy of system with respect to COM is (Kcom)initial=122×13(6−0)2=12J
Let h be the height reached by block.
Using work energy theorem in COM frame: ΔWnet=ΔKE −m1gh=(Kcom)final−(Kcom)initial −2×10×h=0−12 h=122×10 ∴h=0.6m