A stationary bomb explodes into two parts of masses 3kg and 1kg. The total kinetic energy of the two parts after explosion is 2400 J. Then the K.E. of the smaller part is ?

Question

shabbir chatrissa · Answer

the initial momentum is zero. let m 1 =3 kg m 2 =1 kg & their velocities be v 1 & v 2 respectively. according to conservationh of momentum m 1 v 1 + m 2 v 2 = 0 3v 1 + v 2 = 0 v 2 = -3v 1 also, 0.5m 1 v 2 2 + 0.5m 2 v 2 2 = 24000 m 1 v 1 2 + m 2 v 2 2 = 4800 3v 1 2 + v 2 2 = 4800 3v 1 2 +(-3v 1 ) 2 = 4800 12v 1 2 = 4800 v 1 = 20 v 2 = -60 (opp. direction) KE of m 2 is 1800. I suggest you still cross check.

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A stationary bomb explodes into two parts of masses 3kg and 1kg. The total kinetic energy of the two parts after explosion is 2400 J. Then the K.E. of the smaller part is ?

A stationary bomb explodes into two parts of masses 3kg and 1kg. The total kinetic energy of the two parts after explosion is 2400 J. Then the K.E. of the smaller part is ?

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A stationary bomb explodes into two parts of masses 3kg and 1kg. The total kinetic energy of the two parts after explosion is 2400 J. Then the K.E. of the smaller part is ?

A stationary bomb explodes into two parts of masses 3kg and 1kg. The total kinetic energy of the two parts after explosion is 2400 J. Then the K.E. of the smaller part is ?

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