Question

A body of mass 4 kg is moving with a velocity of 4ms⁻¹. Find the ratio of its initial and final Kinetic Energy, if its mass is doubled and velocity is tripled.​

Answer 1

Answer:

Explanation:

case 1

v=4m/s

mass=4kg

k=1/2 mv^2

k= 1/2 * 4 *16

k= 32 J

case 2

on doubling the mass = 8kg

on tripling the velocity = 12m/s

k=1/2 mv^2

k=1/2 *8*12

k=48

so ratio of kinetic energy in both cases would be 32:48

= 2:3

Answer 2

Answer:

$\boxed{\sf Ratio \ of \ initial \ and \ final \ Kinetic \ Energy = 1:18}$

Given:

Initial mass of the body (m) = 4 kg

Initial velocity of the body (v) = 4 m/s

Final mass of the body (m') = Twice the initial mass of the body

Final velocity of the body (v') = Thrice the initial velocity of the body

To Find:

Ratio of initial and final Kinetic Energy

Explanation:

Final mass of the body (m') = 2 × 4

= 8 kg

Final velocity of the body (v') = 3 × 4

= 12 m/s

$\sf KE_i = \frac{1}{2} mv^2$

$\sf KE_f = \frac{1}{2} m'(v')^2$

$\therefore$

$\sf \implies \frac{KE_i}{KE_f} = \frac{ \cancel{\frac{1}{2}} m v^2}{ \cancel{\frac{1}{2}} m'( v' )^2}$

$\sf \implies \frac{KE_i}{KE_f} = \frac{ m v^2}{ m'( v' )^2}$

$\sf \implies \frac{KE_i}{KE_f} = \frac{ 4 \times {4}^{2} }{8 \times {12}^{2} }$

$\sf \implies \frac{KE_i}{KE_f} = \frac{ 4 \times 16}{8 \times 144 }$

$\sf \implies \frac{KE_i}{KE_f} = \frac{4 \times 2}{144}$

$\sf \implies \frac{KE_i}{KE_f} = \frac{2}{36}$

$\sf \implies \frac{KE_i}{KE_f} = \frac{1}{18}$

$\sf \implies KE_i : KE_f=1 :1 8$