Question

A spring ball of mass 0.5 kg is dropped frome some height on falling freely for 10 s it explodes into two fragments of mass ratio 1 :2 the lighter fragment continues to travel downwards with speed of 60 m/s calculate the kinetic energy supplied during the explosion

Answer 1

Answer:

hey mate

Explanation:

here is your answer

given

m = 0.5 kg

g = 10 m/s²

free fall for 10 s

M1 /M2 = 1/2 M1 + M2 = 0.5 kg

v of M1 = 60 m/s

= M1 = m2/2

solution

M1 + M2 = 0.5 kg

$= \frac{m2}{2} + m2 \: = \: 0.5kg$

$\frac{3}{2} m2 = \: 0.5 \: \frac{1}{2}$

M2 = 1/3 kg

$m1 \: = \frac{m2}{2} = \frac{1}{6} kg$

free fall of m : v = GT = 10 × 10 = 100 m/s²

The corresponding KE of m = 1/2 mv²

$= \frac{1}{2} \times 0.5 \: \times {(100)}^{2} = \: \frac{0.5 \: \times 100 \: \times \: 100}{2}$

= 2500 joule

KE of M1 = 1/2 m1v1² = 1/2 × 1/6 × (60)²

$= \frac{60 \times 60}{2 \times 6} = 300j$

by momentum conversation

MV = m1v1 + m2v2

$0.5 \: \times \: 100 \: = \frac{1}{6} \times 60 \: + \frac{1}{3} v2 \: = 10 \: + \frac{ {v}^{2} }{3}$

v²/3 = 50 -10 = 40 V2 = 120 M/S

$ke \: of \:m2 \: = \frac{1}{2} {m2v2}^{2} = \frac{1}{2} \times \frac{1}{3} \times {(120)}^{2}$

$= \frac{120 \times 120}{6} = 20 \times 120 \: = 2400j$

total KE of the fragments = 300 + 2400

= 2700j

KE of the ball just before the explosion = 2500 joule

Hence the KE supplied during the explosion

= 2700 j - 2500 j = 200 j

$thanku$