CBSE Class 11-science Answered
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Asked by suchananag2002 | 19 Sep, 2019, 09:32: PM
Expert Answer
By conservation of momentum, bullet embedded mass Block moves with velocity = (10 × 10-3 × 50√7 ) / ( 500 × 10-3 ) = √7 m/s
Mass block travels along the semi-circular track, till the point B .
At point B, mas block is getting detached from the circular track due to gravitational force mg acting on the block.
Let θ be the angle made by centripetal force with vertical as shown in figure.
Hence, we have, ( m vB2 / R ) = mg cosθ or vB2 = ( R g cosθ ) .......................(1)
By work-energy theorem, (1/2) m vA2 - (1/2) m vB2 = m g (R + R cosθ ) .................(2)
By substituting vA = √7 and vB from eqn.(1) , we get cos θ = 1/2 or θ = 60° from eqn.(2)
If cosθ = 1/2, from eqn.(1), we get vB = 1 m/s.
Once the mass block is detached at B, then it is projected with velocity VB in the direction which is tangent to the circular track at B.
Hence block is projected with projection angle θ = 60° with horizontal.
When the mass block is hitting the horizontal part of track, net vertical distance travelled is ( R + R sinθ ) = 0.2×3/2 = 0.3 m
Time taken t to reach the horizontal part of track, -0.3 = 1×sin60×t -(1/2)gt2 or 10t2 - √3 t - 0.6 = 0 ..........(3)
Accepetable solution of eqn.(3) is t = [√3+ (27)1/2] / 20 = 0.34 s
Horizontal distance travelled during this time t :- vBcosθ × t = 1 × (1/2) × 0.34 = 0.17 m
Horizontal distance of separation point B from point O as shown in figure, ( R - R sinθ ) = 0.2 × [ 1 - (√3/2)]≈ 0.03
Hence horizontal distance of hitting point of the block from point O = 0.17+0.03 = 0.2 m
Hence the block will hit after separation, at the point of junction of horizontal and semi-circular track
Answered by Thiyagarajan K | 21 Sep, 2019, 10:46: AM
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