JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 10)
A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms$$-$$1, the impulse imparted to the gun and velocity of recoil of gun are :
0.2 kg ms$$-$$1, 0.1 ms$$-$$1
0.4 kg ms$$-$$1, 0.05 ms$$-$$1
0.2 kg ms$$-$$1, 0.05 ms$$-$$1
0.4 kg ms$$-$$1, 0.1 ms$$-$$1
Explanation
mBullet = 4g, MGun = 4 kg
vBullet $$ \simeq $$ 50 m/s
Now, PB = Pg
Pg = m $$\times$$ vBullet
= $${4 \over {1000}}$$ $$\times$$ 50
= 0.2 kg m/s
So impulse = 0.2 kg m/s
$${v_G} = {{0.2} \over {{M_{Gun}}}} = {{0.2} \over 4} = 0.05$$ m/s
vBullet $$ \simeq $$ 50 m/s
Now, PB = Pg
Pg = m $$\times$$ vBullet
= $${4 \over {1000}}$$ $$\times$$ 50
= 0.2 kg m/s
So impulse = 0.2 kg m/s
$${v_G} = {{0.2} \over {{M_{Gun}}}} = {{0.2} \over 4} = 0.05$$ m/s
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