Q.32. In the given figure shown a small block B of mass m is released from the top of the smooth movable wedge A of the same mass m. The height of wedge A shown in figure is h = 100 cm. B ascends movable smooth wedge C of the same mass. Neglecting friction any where find the maximum height (in cm) attained by block B on wedge C.


(1) 25
(2) 30
(3) 35
(4) 40

by conservation of momentum for block and wedge B(along x direction only),
block B and wedge will have same velocity oppositely,
so, energy conservation from top to bottom,
loss of PE = gain in total KE
mgh = 1/2.(2m)v2 
v2 = 100g
 
so, we found the velocity of block when block reaches down
now, again using law of conservation of momentum for block and wedge C.
both block and wedge C will have same velocity(along x) when block reached on topmost point(say v1),
initial momentum = final toatal momentum
mv = mv1(wedege C) + mv1(block)
v1 = v/2

now again by energy conservation from bottom to top
1/2mv​2  = 1/2 m (v1​)2 (for wedge C)​​ + 1/2 m (v1)​2 (for bblock) ​​+ mgH (block and wege C has same velocity v1 at topmost point)
 by solving this, H = 25 m

  • 12
remember to put v1 = v/2 in last equation and put v2 as 100g as i solved above
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