Work, Energy and Power
- A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is :
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Let the initial velocity of the shell be v, then by the conservation of momentum mv = Mv'
where v' = velocity of gun.∴ v' = m v M Now, total K.E. = 1 mv2 + 1 Mv'2 2 2 = 1 mv2 + 1 M m 2 v2 2 2 M = 1 mv2 1 + m 2 M = 1 × 0.2 1 + 0.2 v2 = (0.1 × 1.05)v2 2 4
But total K.E. = 1.05 kJ = 1.05 × 103 J
∴ 1.05 × 103 = 0.1 × 1.05 × v2v2 = 1.05 × 103 = 104 0.1 × 1.05
∴ v = 102 = 100 ms–1.Correct Option: A
Let the initial velocity of the shell be v, then by the conservation of momentum mv = Mv'
where v' = velocity of gun.∴ v' = m v M Now, total K.E. = 1 mv2 + 1 Mv'2 2 2 = 1 mv2 + 1 M m 2 v2 2 2 M = 1 mv2 1 + m 2 M = 1 × 0.2 1 + 0.2 v2 = (0.1 × 1.05)v2 2 4
But total K.E. = 1.05 kJ = 1.05 × 103 J
∴ 1.05 × 103 = 0.1 × 1.05 × v2v2 = 1.05 × 103 = 104 0.1 × 1.05
∴ v = 102 = 100 ms–1.