Question

A 0.40 kg cart is moving at 3 m/s on a frictionless track where it collides with a stationary 0.80 kg cart and two stick together. What is the final velocity of the pair ?

Answer #1

Masses of carts m_{1} = 0.4 kg

m_{2} = 0.8 kg

Initial Velocities of carts

V_{1} = 3 m/s

V_{2} = 0 (i.e, it is at rest)

Final velocity of the pair V = ?

From law of conservation of momentum,

m_{1} v_{1} + m_{2} v_{2} = (m1
+ m2)V

m_{1} v_{1} + 0 = (m_{1} +
m_{2})V

V = m_{1} v_{1}/m_{1} +
m_{2}

= (0.4) (3)/ 0.4 + 0.8 = 1m1s

m_{1} v_{1} + m_{2} v_{2} = (m1
+ m2)V

m_{1} v_{1} + 0 = (m_{1} +
m_{2})V

V = m_{1} v_{1}/m_{1} +
m_{2}

= (0.4) (3)/ 0.4 + 0.8 = 1m1s

m_{1} v_{1} + m_{2} v_{2} = (m1
+ m2)V

m_{1} v_{1} + 0 = (m_{1} +
m_{2})V

V = m_{1} v_{1}/m_{1} +
m_{2}

= (0.4) (3)/ 0.4 + 0.8 = 1m1s

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