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Standard XII

Physics

Perfectly Elastic Collision

The car A o...

Question

The car $A$ of mass $1500 k g$ travelling at $40 m / s$ collides with another car $^{'} B^{'}$ of mass $1250 k g$ travelling at $25 m / s$ in the same direction. After collision the velocity of car $^{'} A^{'}$ becomes $30 m / s$ . calculate the velocity of car $^{'} B^{'}$ after the collision.

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Solution

BY LAW OF CONSERVATION OF MOMENTUM WE HAVE

1500 * 40 + 1250* 25 = 30 * 1500 + 1250* Vb
60000 + 31250 = 45000 + 1250*Vb
91250 - 45000 = 1250Vb
46250 = 1250Vb
therefore Vb = 46250 / 1250
= 37 m/sec

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The car A of mass 1500kg travelling at 40m/s collides with another car ′B′ of mass 1250kg travelling at 25m/s in the same direction. After collision the velocity of car ′A′ becomes 30m/s. calculate the velocity of car ′B′ after the collision.

BY LAW OF CONSERVATION OF MOMENTUM WE HAVE 1500 * 40 + 1250* 25 = 30 * 1500 + 1250* Vb60000 + 31250 = 45000 + 1250*Vb91250 - 45000 = 1250Vb46250 = 1250Vbtherefore Vb = 46250 / 1250 = 37 m/sec

The car $A$ of mass $1500 k g$ travelling at $40 m / s$ collides with another car $^{'} B^{'}$ of mass $1250 k g$ travelling at $25 m / s$ in the same direction. After collision the velocity of car $^{'} A^{'}$ becomes $30 m / s$ . calculate the velocity of car $^{'} B^{'}$ after the collision.

BY LAW OF CONSERVATION OF MOMENTUM WE HAVE

1500 * 40 + 1250* 25 = 30 * 1500 + 1250* Vb
60000 + 31250 = 45000 + 1250*Vb
91250 - 45000 = 1250Vb
46250 = 1250Vb
therefore Vb = 46250 / 1250
= 37 m/sec