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A body of mass M at rest explodes in three pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3m/s and

4m/s respectively. The third will thrown off with a velocity of ?

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Initial mass = M initial velocity = 0

After explosion, m1 = M/4 v1 = 3m/s i (i means along x-direction) m2 = M/4 v2 = 4m/s j (j means along y-direction) m3 = M-(M/4 + M/4) = M/2 v3 = V

As there is no external force, momentum before explosion = momentum after explosion ⇒M(0) = m1v1 + m2v2 + m3v3 ⇒ 0 = M/4 (3)i + M/4 (4)j + M/2(V) ⇒ 0 = M [ (3/4)i + 1j + V/2 ] ⇒ 0 = (3/4)i + 1j + V/2 ⇒ V/2 = -[(3/4)i + 1j] ⇒ V = 2×-[(3/4)i + 1j] ⇒ V = -[ (3/2)i +(2)j ] ⇒ V = -(3/2)i -2j |V| = √[(3/2)² + 2²] = 2.5m/s

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Momentum is a vector. Initial momentum of M is zero. So resultant of vector sum of all three momenta of 3 pieces is zero. Let us find the resultant of the momenta of two masses M/4.

Momenta are 3 M/4 and 4M/4 = M. The angle between them is 90⁰.

So θ = 36.87⁰.

So the momentum of the mass M/2 = 1.25 M.

Its velocity = 1.25 M / (M/2) = 2.5 m/sec Its direction is 126.87⁰ from direction of M/4 having velocity 3 m/s and 143.13⁰ from the direction of M/4 having velocity 4 m/s