Given, the Bullet’s mass (m1) = 50 g
The rifle’s mass (m2) = 4 kg = 4000g
Initial velocity of the fired bullet (v1) = 35 m/s
Let the recoil
velocity be v2.
Since the rifle was initially at rest, the initial momentum of the rifle = 0
Total momentum of the rifle and bullet after firing = m1v1 + m2v2
As per the law of conservation of momentum, the total momentum of the rifle and the bullet after firing = 0 (same as initial momentum)
Therefore, m1v1 + m2v2 = 0
\(This \,implies \,that\, v_2 = -\frac{m_1v_1}{m_2}\)
\( = -\frac{50g\,\times \, 35ms^{-1}}{4000g}\)
= - 0.4375 m/s
Therefore, the recoil velocity of the rifle is 0.4375 meters per second in the direction opposite to the trajectory of the bullet (backward direction).
Answered by Vishal kumar | 2 years agoA motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required
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The following is the distance-time table of an object in motion:
| Time (seconds) | Distance (meters) |
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 84 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
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