Question:
A machine gun has a mass 5 kg. It fires 50 gram bullets at the rate of 30 bullets per minute at a speed of 400 m/s. What force is required to keep the gun in position?
A machine gun has a mass 5 kg. It fires 50 gram bullets at the rate of 30 bullets per minute at a speed of 400 m/s. What force is required to keep the gun in position?
Show Hint
The force needed to keep a machine gun in position can be found using the rate of change of momentum for each bullet fired.
Updated On: Jan 12, 2026
- 10 N
- 5 N
- 15 N
- 30 N
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: The force required to keep the machine gun in position is related to the rate of change of momentum. The rate of change of momentum is the force exerted on the gun due to the bullets being fired.
Step 2: The momentum of each bullet is given by \( p = mv \), where \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) and \( v = 400 \, \text{m/s} \). The rate of change of momentum is: \[ F = \text{rate of change of momentum} = \dfrac{\Delta p}{\Delta t} = \left( 0.05 \times 400 \right) \times \dfrac{30}{60} = 15 \, \text{N}. \]
Final Answer: \[ \boxed{15 \, \text{N}} \]
Step 2: The momentum of each bullet is given by \( p = mv \), where \( m = 50 \, \text{g} = 0.05 \, \text{kg} \) and \( v = 400 \, \text{m/s} \). The rate of change of momentum is: \[ F = \text{rate of change of momentum} = \dfrac{\Delta p}{\Delta t} = \left( 0.05 \times 400 \right) \times \dfrac{30}{60} = 15 \, \text{N}. \]
Final Answer: \[ \boxed{15 \, \text{N}} \]
Was this answer helpful?
0
0
Top Questions on Mechanics
- Match List-I with List-II
\[ \begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A) \; P \text{ and } Q \text{ are two perpendicular forces, acting at a point} & (I) \; R = |P - Q| \\ (B) \; P \text{ and } Q \text{ are equal, forces acting at a point at an angle } \alpha & (II) \; R = P + Q \\ (C) \; P \text{ and } Q \text{ are acting at a point in same direction} & (III) \; R = 2P \cos(\tfrac{\alpha}{2}) \\ (D) \; P \text{ and } Q \text{ are acting at a point in opposite direction} & (IV) \; R = \sqrt{P^2 + Q^2} \\ \hline \end{array} \]
Choose the correct answer from the options given below:
- Two forces acting at a point of a body are equilibrium if and only if they
(A) are equal in magnitude
(B) have same direction
(C) have opposite direction
(D) act along the same straight line
(E) are not equal in magnitude but have same direction
Choose the correct answer from the options given below: - If three forces of magnitudes 8 newtons, 5 newtons and 4 newtons acting a point are in equilibrium, then the angle between the two smaller forces is
- If two stones are thrown vertically upwards with their velocities in the ratio 2:5, then the ratio of the maximum heights attained by the stones is
- In a perfectly elastic collision if m₁ and m₂ be the masses and v₁ and v₂ be the velocities of two colliding particles. Then velocity after the collision, if particles stick together will be:
View More Questions
Questions Asked in VITEEE exam
- In a code language, 'TIGER' is written as 'JUISF'. How will 'EQUAL' be written in that language?
- VITEEE - 2025
- Odd one Out
- Find the odd one out from the following series: 5, 10, 20, 40, 100, 150, 200
- VITEEE - 2025
- Odd one Out
- How many numbers between 0 and 9 look the same when observed in a mirror?
- VITEEE - 2025
- Odd one Out
- Find the value of \( x \) in the equation \( 4x - 2 = 10 \).
- VITEEE - 2025
- Algebra
- MNO : PQR :: XYZ : ?
- VITEEE - 2025
- Odd one Out
View More Questions