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प्रश्न
Solve the following problem.
A spring ball of mass 0.5 kg is dropped from some height. On falling freely for 10 s, it explodes into two fragments of mass ratio 1:2. The lighter fragment continues to travel downwards with a speed of 60 m/s. Calculate the kinetic energy supplied during the explosion.
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उत्तर
m1 + m2 = 0.5 kg, m1 : m2 = 1 : 2,
m1 = `1/6` kg
∴ m2 = `1/3` kg
Initially, when the ball is falling freely for 10s,
v = u + at = 0 + 10(10)
∴ v = 100 m/s = u1 = u2
(m1 + m2)v = m1v1 + m2v2
∴ `0.5 xx 100 = 1/6(60) + 1/3 "v"_2`
∴ 50 = 10 + `1/3 "v"_2`
∴ 40 = `1/3 "v"_2`
∴ v2 = 120 m/s
∴ Δ K.E. = `1/2 "m"_1"v"_1^2 + 1/2"m"_2"v"_2^2 - 1/2 ("m"_1 + "m"_2)"u"^2`
∴ Δ K.E. = `1/2(1/6) xx 60^2 + 1/2 xx 1/3 xx (120)^2 - 1/2 xx 0.5 xx (100)^2`
= 300 + 2400 - 2500
∴ K.E. = 200 J
Kinetic energy supplied is 200 J.
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