A bomb of mass $3.0 kg$ explodes in air into two pieces of masses $2.0 kg$ and $1.0 kg$ . The smaller mass goes at a speed of $80 m/s$ . The total energy imparted to the two fragments is:

1.07 kJ

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2.14 kJ

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2.4 kJ

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4.8 kJ

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Solution

The correct option is D $4.8 kJ$

Applying conservation of momentum

$2 \times v = 1 \times 80$

$v = 40 m/s$

Total energy of pieces

$= \frac{1}{2} \times 2 \times 40 \times 40 + \frac{1}{2} \times 1 \times 80 \times 80$

$= 1600 + 3200 = 4800 J = 4.8 kJ$

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A bomb of mass 3.0kg explodes in air into two pieces of masses 2.0kg and 1.0kg. The smaller mass goes at a speed of 80m/s. The total energy imparted to the two fragments is:

The correct option is D 4.8kJ Applying conservation of momentum 2×v=1×80 v=40m/s Total energy of pieces =12×2×40×40+12×1×80×80 =1600+3200=4800J=4.8kJ

A bomb of mass $3.0 kg$ explodes in air into two pieces of masses $2.0 kg$ and $1.0 kg$ . The smaller mass goes at a speed of $80 m/s$ . The total energy imparted to the two fragments is:

The correct option is D $4.8 kJ$

Applying conservation of momentum

$2 \times v = 1 \times 80$

$v = 40 m/s$

Total energy of pieces

$= \frac{1}{2} \times 2 \times 40 \times 40 + \frac{1}{2} \times 1 \times 80 \times 80$

$= 1600 + 3200 = 4800 J = 4.8 kJ$