Question

Two objects of masses 1 kg and 4 kg are at rest next to a compressed spring. The spring stores an energy j. When it is released, it gives all its energy to the blocks and falls to the ground. Now, their speeds are vi v4. Find v4. (There is no friction.)

Answer 1

Solution :

Given :

m₁ = 1 kg

m₄ = 4 kg

U_Spring = 2250 J

.

According to the conservation of energy :

∴ KE₁ + KE₄ = U_Spring

∴ (1/2) m₁ (v₁)² + (1/2) m₄ (v₄)² = U_Spring

∴ (v₁)² + 4 (v₄)² = 4500 J . . . (i)

.

Now, According to the conservation of momentum : P_f = P_i

∴ m₁ v₁ + m₄ v₄ = 0

∴ (1 kg) v₁ + (4 kg) v₄ = 0

∴ v₁ = - 4 v₄

_.

Therefore, From equation (i) :

∴ (- 4 v₄)² + 4 (v₄)² = 4500 J

∴ 16 (v₄)² + 4 (v₄)² = 4500 J

∴ 20 (v₄)² = 4500 J

∴ (v₄)² = 225 J

∴ v₄ = 15 m/s