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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 | 20 Feb, 2020, 11:24: AM
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/s2

Hence distance S travelled by cannon while firing bullet is obtained from ,  v2 = u2 - 2aS

where v is final speed wich is zero here, u is inital speed and a is retardation

Hence S = u2 / (2a) = ( 0.4 × 0.4 ) / ( 2 × .98 ) = 0.081 m  = 8.1 cm
Answered by Thiyagarajan K | 22 Feb, 2020, 11:18: AM

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