Question

A block of mass 1 kg
moving with a speed of
4 ms' collides with another
block of mass 2 kg which
is at rest. The lighter block
comes to rest after
collision. The loss in KE of
the system is​

Answer 1

Answer:

Loss in K.E of the system = 4 J

Explanation:

Given:

Mass of the first block = 1 kg

Mass of the second block = 2 kg

Initial velocity of the first block = 4 m/s

Initial velocity of the second block = 0 m/s

Final velocity of the first block = 0 m/s

To Find:

Loss in kinetic energy of the system

Solution:

First find the final velocity of the second block,

By conservation of momentum we know that,

Initial momentum = Final momentum

Hence,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where

m₁ = mass of first block

m₂ = mass of second block

u₁ = initial velocity of first block

u₂ = initial velocity of second block

v₁ = final velocity of first block

v₂ = final velocity of second block

Substitute the data,

1 × 4 + 2 × 0 = 1 × 0 + 2 × v₂

4 = 2 v₂

v₂ = 4/2

v₂ = 2 m/s

Hence the final velocity of the second block is 2 m/s.

Now,

Loss of K.E is given by,

$\sf \Delta K.E=Initial\:K.E(K_i)-Final\:K.E(K_f)$

Finding the initial Kinetic energy,

K.E = 1/2 × m₁ × (u₁)²

K.E = 1/2 × 1 × 16

K.E = 8 J

Now finding final K.E,

K.E = 1/2 × m₂(v₂)²

K.E = 1/2 × 2 × 4

K.E = 4 J

Substitute the value,

ΔK.E = 8 - 4

ΔK.E = 4 J

Hence the loss in K.E of the system is 4 J.