Mechanics Momentum

a. By the law of conservation of momentum

m1u = (m1 + m2)v

(0.02)(500) = (3 + 0.02)v

Velocity of the block after the bullet is embedded in it, v = 3.31 ms-1


b. Loss of K.E. = Work done against friction

1/2 (m1 + m2)v2 = fs

1/2 (3 + 0.02)(3.31)2 = 6s

Total distance travelled, s = 2.76 m


c.i. It is the same as the answer in (a). It is 3.31 ms-1.

ii. By Newton's 2nd law of motion,

F = (m1 + m2)a

6 = (3 + 0.02)a

Deceleration of the block, a = 1.99 ms-2

Now, the friction acting on the trolley by the block is 6N also.

By Newton's 2nd law of motion,

F = ma

6 = 5a

Acceleration of the trolley, a = 1.2 ms-2


iii. For the block to stop moving along the trolley, they should have the same linear speed (The relative velocity between them is zero).

The momentum of the system is conserved since there is no net force acting on the system.

By the law of conservation of momentum,

Initial momentum of the bullet = Final momentum of the bullet, trolley and block

(0.02)(500) = (0.02 + 3 + 5)v

Final speed, v = 1.25 ms-1

Consider the block, by equation of motion,

v = u + at

1.25 = 3.31 + (-1.99)t

Time taken, t = 1.04 s


iv. Consider the block,

By equation of motion,

v2 = u2 + 2as

(1.25)2 = (3.31)2 + 2(-1.99)s

Distance, s = 2.36 m

But this is the distance the block moves relative to the ground.

Consider the trolley, by s' = ut + 1/2 at2

s' = 0 + 1/2 (1.2)(1.04)2

s' = 0.65 m

So, the required distance = 2.36 - 0.65 = 1.71 m