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A block of mass 1 kg is moving towards a...

A block of mass 1 kg is moving towards a movable wedge of mass 2 kg as shown in figure. All surfaces are smooth. When the block leaves the wedge from top, its velocity is making an angle `theta =30^(@)` with horizontal.

The value of `v_(0)` in m/s is

A

1.9 m

B

2.7 m

C

1.6 m

D

1.45 m

Text Solution

Verified by Experts

The correct Answer is:
C
`H = 1.45 + (v_b^2 sin^2 30^@)/(2g) " " = 1.45 + (3v^2)/(8g) " "(v_b = sqrt(3)v)`
`= 1.45 + (3v_0^2)/(98 g) " "(v = 2/7 v_0) " " = 1.45 + (3 xx 49)/(98 xx 10) " " = 1.6 m`
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