Question

a spring ball of mass 0.5 kg is dropped from the some height. on falling freely for 10s,it explodes into two fragments of mass ratio 1:2. the lighter fragment continues to travel downward with speed of 60m/s. calculate the K.E supplied during explosion.​

Answer 1

Answer:

Given

Mass:0.5 kg

Mass lighter fragment: 5/30

Velocity: 60m/s

Explanation:

As we know,

K. E=1/2.m.v^2

=1/2(5/30){(60)(60)}

=300joule

Therefore, the kinetic energy of the lighter fragment body is 300joule

Answer 2

The kinetic energy supplied during explosion is 180.5 J

Explanation:

Given that,

Mass of spring ball = 0.5 kg

Time = 10 s

Mass ratio after explosion =1:2

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s=\dfrac{1}{2}\times9.8\times(10)^2$

$s=490\ m$

We need to calculate the velocity of ball before explosion

Using formula of energy conservation

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

Put the value into the formula

$v=\sqrt{2\times9.8\times490}$

$v=98\ m/s$

We need to calculate the velocity after explosion

Using conservation of momentum

$mv=m_{1}v_{1}+m_{2}v_{2}$

$0.5\times98=\dfrac{0.5}{3}\times60+\dfrac{2}{3}\times0.5\times v_{2}$

$v_{2}=\dfrac{39\times3}{2\times0.5}$

$v_{2}=117\ m/s$

We need to calculate the kinetic energy supplied during explosion

Using formula of kinetic energy

Kinetic energy supplied during explosion = Final kinetic energy of fragment - Initial kinetic energy of ball

$K.E=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2-\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times \dfrac{0.5}{3}\times60^2+\dfrac{1}{2}\times \dfrac{2\times0.5}{3}\times117^2-\dfrac{1}{2}\times0.5\times98^2$

$K.E=180.5\ J$

Hence, The kinetic energy supplied during explosion is 180.5 J

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