A) \[0,-\,150\text{ }m/s\]
B) \[150\text{ }m/s,-\,50\text{ }m/s\]
C) 100 m/s, 150 m/s
D) 100 m/s, 50 m/s
Correct Answer: A
Solution :
[A]Let the first truck move to the right and the second to the left. Let us take the rightward direction as positive. In the horizontal direction, friction being absent, no external force is acting on the system. Hence momentum is conserved. |
The momentum carried away by a sack thrown from the first is \[m{{v}_{1}}.\] the momentum brought in the second sack to the first is \[m(-{{v}_{2}}).\] |
\[\therefore \] By the law of conservation of momentum of (truck +sacks of rice) |
\[(M+m)\,{{v}_{1}}-m{{v}_{1}}+m\,(-{{v}_{2}})=(M+m)\,{{v}_{1}}'\] |
Where \[{{v}_{1}}'\] is the new velocity of truck 1. |
\[\therefore M{{v}_{1}}-m{{v}_{2}}=(M+m)\,{{v}_{1}}'\] |
\[\therefore \,\,{{v}_{1}}'=\frac{M}{M+m}{{v}_{1}}-\frac{m}{M+m}{{v}_{2}}\] |
\[=\frac{200}{250}.50-\frac{50}{250}.200=0\] |
Similarly considering conservation of momentum of the second (truck \[\div \] sacks of rice), we get for the second truck. |
\[{{v}_{2}}'=\left[ \frac{M}{M+m}{{v}_{2}}-\frac{m}{M+m}.{{v}_{1}} \right]\] |
\[=-\,\left[ \frac{200}{250}\times 200-\frac{50}{250}\times 50 \right]=-\,150\,\,m/s.\] |
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