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Standard XII

Physics

Defining Kinetic Energy

A bomb of mas...

Question

A bomb of mass $16$ kg at rest explodes into two pieces of masses of $4$ kg and $12$ kg. The velocity of the $12$ kg mass is $4$ m/s. The kinetic energy of the other mass is

96

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144

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288

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192

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Solution

The correct option is C $288$ J

$By using the law of conservation of momentum, we get,$
$M u_{1} = m_{1} V_{1} + m_{2} V_{2}$
$12 \times 4 + 4 \times V_{2} \Rightarrow V_{2} = - 12 m/s$
$K E = \frac{1}{2} m_{2} V_{2}^{2} = \frac{1}{2} \times 4 \times 144 = 288 J$

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A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The velocity of the 12 kg mass is 4 m/s. The kinetic energy of the other mass is

The correct option is C 288 J By using the law of conservation of momentum, we get, Mu1=m1V1+m2V212×4+4×V2⇒V2=−12 m/sKE=12m2V22=12×4×144=288 J

A bomb of mass $16$ kg at rest explodes into two pieces of masses of $4$ kg and $12$ kg. The velocity of the $12$ kg mass is $4$ m/s. The kinetic energy of the other mass is

The correct option is C $288$ J

$By using the law of conservation of momentum, we get,$
$M u_{1} = m_{1} V_{1} + m_{2} V_{2}$
$12 \times 4 + 4 \times V_{2} \Rightarrow V_{2} = - 12 m/s$
$K E = \frac{1}{2} m_{2} V_{2}^{2} = \frac{1}{2} \times 4 \times 144 = 288 J$