The mass of a cannon is 500 kg. From this cannon a bullet of 10 kg mass was thrown at a speed of 20 m / s. How long does a cannon retreat if the resistance of the land is one tenth the weight of the cannon?

### Asked by asishsengupta51 | 20th Feb, 2020, 11:24: AM

Expert Answer:

### By conservation of momentum, we get initial speed of cannon as , M V = - m v
where M = 500 kg is mass of canon, V is its initial speed , m is mass of bullet = 10 kg and v = 20 m/s is speed of bullet.
we get V = - 10 × 20 / 500 = - 0.4 m/s
if the resistance given by land is (1/10)th weight, i.e. resistive force = (1/10) × 500 × 9.8 N = 490 N
Retardation = force/mass = 490/500 = 0.98 m/s^{2}
Hence distance S travelled by cannon while firing bullet is obtained from , v^{2} = u^{2} - 2aS
where v is final speed wich is zero here, u is inital speed and a is retardation
Hence S = u^{2} / (2a) = ( 0.4 × 0.4 ) / ( 2 × .98 ) = 0.081 m = 8.1 cm

^{2}

^{2}= u

^{2}- 2aS

^{2}/ (2a) = ( 0.4 × 0.4 ) / ( 2 × .98 ) = 0.081 m = 8.1 cm

### Answered by Thiyagarajan K | 22nd Feb, 2020, 11:18: AM

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