CBSE Class 11-science Answered
A body of mass 1 kg is initially at rest, explodes and breakers into three fragments of masses in the ratio 1:1:3. The two fragments of equal masses fly off perpendicular to each other with a speed of 30m/s each. What is the velocity of the heavier fragment ?
Ans. 14.1m/s, at an angle of 135° relative to the second fragment
Asked by rrpapatel | 18 Oct, 2018, 04:04: PM
Expert Answer
Initial mass 1 kg breaks up into three fragments in the ratio 1:1:3 means,
equal fragments are of mass 0.2 kg and other one is 0.6 kg.
Since it is given that equal fragments are moving in a direction perpendicular to each other,
let us assume they are moving along X any Y axis of cartesian coordinate system.
Also let us assume 0.6 kg mass moving with speed v m/s along the direction that makes an angle α with X-axis as shown in figure.
Momentum conservation equation along X-direction :- 0.2×30 +0.2×0+0.6×v cosα = 0 or v cosα = -10 ...........................(1)
Momentum conservation equation along Y-direction :- 0.2×0 +0.2×30+0.6×v sinα = 0 or v sinα = -10 ...........................(2)
If we solve (1) and (2) , we get v = 10√2 m/s and sinα = cosα = -1/√2 , hence α = 225°
Hence if two equal fragments of mass 0.2 kg are moving with speed 30 m/s respectively in positive X-direction and positive Y-direction,
then the third fragment of mass 0.6 kg is moving with speed 10√2 m/s in third quadrant making an angle 225º with +ve X-axis.
Answered by Thiyagarajan K | 18 Oct, 2018, 09:25: PM
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