Question

On a frictionless air track, a 0.150 kg glider moving at 1.20 m/s to the right collides with and sticks to a stationary 0.250 kg glider. A) What is the momentum of this two glider system before the collision? B) What must be the net momentum of this system after the collision? Why? C) Use answers from a and b to find the speed of the gliders after the collision. D) Is kinetic energy conserved during the collision?

Answer #1

A) P_{total initial}= P_{1i}+P_{2i} =
m_{1}v_{1i}+m_{2}v_{2i} =
0.150*1.20+0.250*0 = 0.18 kg*m/s

B) Net momentum after the collision but same same as net momentum before coliision

Hence,

P_{total final} = P_{total initial}

C) Since two glider sticks together,

v_{1f}=v_{2f} = v_{common}

So,

P_{total initial} =P_{total final}

P_{1i}+P_{2i} =
P_{1f}+P_{2f}

m_{1}v_{1i}+m_{2}v_{2i} =
m_{1}v_{1f}+m_{2}v_{2f}

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

0.18= (0.150+0.250)
v_{common}
=> **v _{common} = 0.45 m/s**

D) KEi= 1/2m1v1i^{2}+1/2v2i^{2} =
½*0.150*1.2^{2}+½*0.250*0^{2} = 0.108 J

KEf= 1/2m1v1i^{2}+1/2v2i^{2} =
½*0.150*0.45^{2}+½*0.250*0.45^{2} =0.0405 J

Since KEi is not equal to KEf kinetic energy not conserve here.

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