Question

A) A cannon of mass m, 5000 kgs located on a smooth horizontal platform
fires a shell of mass m, 100 kg in horizontal direction with a velocity
v=150 m/s. Find the velocity of the cannon after it is shot.
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Answer 1

Answer:

m1=5000kg

m2=100kg

u1=150m per second

u2=0

v1=v

v2=v

total momentum before collision =total momentum after collision

m1u1+m2u2=m1v1+m2v2

5000×150+100×0=5000×v+100×v

750000=5100v

v=750000/5100

v=7500/51

Answer 2

The velocity of the cannon after it is shot is -3 m/s.

Explanation:

Given mass of cannon, M = 5000 kg

Initial velocity of cannon, V = 0

Mass of shell, m = 100 kg

Velocity of shell, v = 150 m/s

The velocity of the cannon after it is shot, u = ?

To calculate the velocity of the cannon, use the conservation of momentum.

 The momentum of the system before shot = momentum of the system after shot.

$M\times V=M\times u+m\times v$

Substitute the given values, we get

$5000 kg\times 0=5000 kg\times u+100 kg\times 150 m/s$

$u=-\frac{15000}{5000} =-3 m/s$.

Negative sign shows cannon move backward direction when it is shot.

Thus, the velocity of the cannon after it is shot is -3 m/s.

Topic : Conservation of momentum

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