Question

Two blocks A and B of masses 30 kg and 20 kg are approaching towards each other at right angles with velocities 1 m/s and 2
m/s respectively on smooth surface. They collide and stick together. The loss of energy is​

Answer 1

Answer:

nice

Explanation:

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Answer 2

Answer:

ΔKE = 1 J

Explanation:

given,

Mass of block A (M)= 30 Kg

speed, v₁ = 1 m/s

Mass of block B (m)= 20 Kg

speed, v₂ = 2 m/s

using conservation of momentum

M v₁ + m v₂ = (M + m) V

30 x 1 + 20 x 2 = (30+20)V

50 V = 70

 V = 1.4 m/s

initial kinetic energy

$KE_1 = \dfrac{1}{2}Mv_1^2 + \dfrac{1}{2}mv_2^2$

$KE_1 = \dfrac{1}{2}\times 20 \times 1^2 + \dfrac{1}{2}\times 20 \times 2^2$

KE₁ = 50 J

final kinetic energy

$KE_2= \dfrac{1}{2}(M+m)V^2$

$KE_2= \dfrac{1}{2}\times 50 \times 1.4^2$

KE₂ = 49 J

Loss in kinetic energy

ΔKE = KE₁ - KE₂

ΔKE = 50 - 49

ΔKE = 1 J

Loss in kinetic energy is equal to 1 J