A machine gun is mounted on a 2000 ,kg car on | vedclass

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(B) The thrust force $F$ exerted on the system is given by the rate of change of momentum of the bullets fired.$F = v_{rel} \cdot \frac{dm}{dt}$Here,$

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) The thrust force $F$ exerted on the system is given by the rate of change of momentum of the bullets fired.$F = v_{rel} \cdot \frac{dm}{dt}$Here,$v_{rel} = 500 \,m/s$ is the velocity of the bullet with respect to the car.The mass of one bullet is $m_b = 10 \,g = 0.01 \,kg$.The number of bullets fired per second is $n = 10 \,bullets/s$.Therefore,the rate of mass ejection is $\frac{dm}{dt} = n 	imes m_b = 10 	imes 0.01 \,kg/s = 0.1 \,kg/s$.Substituting these values into the thrust formula:$F = 500 \,m/s 	imes 0.1 \,kg/s = 50 \,N$.Thus,the average thrust on the system is $50 \,N$.

$A$ machine gun is mounted on a $2000 \,kg$ car on a horizontal frictionless surface. At some instant,the gun fires bullets of mass $10 \,g$ with a velocity of $500 \,m/s$ with respect to the car. The number of bullets fired per second is $10$. The average thrust on the system is ........ $N$.

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