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# Test: Conservation Of Linear Momentum

## 10 Questions MCQ Test Science Class 9 | Test: Conservation Of Linear Momentum

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This mock test of Test: Conservation Of Linear Momentum for Class 9 helps you for every Class 9 entrance exam. This contains 10 Multiple Choice Questions for Class 9 Test: Conservation Of Linear Momentum (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Conservation Of Linear Momentum quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this Test: Conservation Of Linear Momentum exercise for a better result in the exam. You can find other Test: Conservation Of Linear Momentum extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
QUESTION: 1

### Rockets work on the principle of conservation of:

Solution:
Rocket works on the conservation of momentum. In a rocket, the fuel burns and produces gas at high temperature. These gases are ejected out of the rocket from a nozzle at the back side of the rocket. The ejecting gas exerts a forward force on the rocket which help in accelerating. Through the mass of gases escaping per second is very small and their momentum is very large due to their tremendous velocity of escape. An equal and opposite momentum is imparted to the rocket which despite its large mass builds up a high velocity.
QUESTION: 2

### The law of conservation of momentum holds for:

Solution: Conservation of momentum. A conservation law stating that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.
QUESTION: 3

### A gun of mass 5kg fires 50gm bullet with a velocity of 200m/s. Find the recoils velocity of the gun.

Solution:

M of bullet = 50g= 50/1000 = 0.05kg
u= 200m/s , v=0m/s
m of gun = 5kg
u= 0m/s  , v= ?
p of bullet before firing = 0.05x200 = 10kgm/s
p of gun before firing = 5*0 = 0kgm/s
Total p before firing = 10 + 0 = 10kgm/s
p of bullet after firing = 0.05x0 = 0kgm/s
let recoil velocity of gun = v m/s
p of gun after firing = 5v kgm/s
total p after firing = 5v + 0 = 5v kgm/s
According to law of conservation of momentum(p)
total p before firing = total p after firing
10    =   5v
10/5    = v
2     =   v
∴ recoi v of gun = - 2 m/s

QUESTION: 4

Which of the following implies the greatest precision?

Solution: Accelaration =(v-u)/t a=30-20/2sec=5m/s f = ma. F = 2kg*5m/s =10N
QUESTION: 5

The law of conservation of momentum states that the sum of momenta of two objects before collision is equal to the sum of momenta after the collision provided that ______

Solution:

The law of conservation of momentum states that for two objects colliding in an isolated system, the total momentum before and after the collision is equal. This is because the momentum lost by one object is equal to the momentum gained by the other.

QUESTION: 6

The momentum of a body of given mass is proportional to:

Solution:
QUESTION: 7

A ball of mass 5 kg moving at 3 m/s collides with another ball of mass 3 kg moving at 5 m/s in the same direction. If the balls move together after the collision in the same direction, find their common velocity.​

Solution: Let 'v' be the common velocity. . Then Using law of conservation of momentum... we have: m1u1+m2u2=(m1+m2)v... =》(5×3)+(3×5)=(5+3)v.. =》30=8v.. =》v=30/8=3.75 m/s..
QUESTION: 8

Two objects each of mass 1.5 Kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5m/s before the collision during which they stick together. What will be the velocity of the combined object after the collision?​

Solution:
Mass ofone of the objects,m1= 1.5 kg

Mass of the other object,m2= 1.5 kg

Velocity ofm1before collision,v1= 2.5 m/s

Velocity ofm2, moving in opposite direction before collision,v2= 2.5' m/s

(Negative sign arises because massm2is moving in an opposite direction)

After collision, the two objects stick together.

Total mass of the combined object =m1+m2

Velocity of the combined object =v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1v1+m2v1= (m1+m2)v

1.5(2.5) + 1.5 (2.5') = (1.5 + 1.5)v

3.75' + 3.75 = 3v

v= 0

Hence, the velocity of the combined object after collision is 0 m/s.
QUESTION: 9

A machine gun of mass 24 kg fires 20g bullets at the rate of 8 bullets per second, with a speed of 250 m/s. The force required to hold the gun in position is:

Solution:
In the given problem,

mg = 24 kg ; vb = 250 m/s

mass of the bullet = 20g = 20 x 10^-3 kg

no. of bullets fired/sec = 8
F = Chnage in momentum/time = mb*vb - mb*ub/t

For t =  1 sec
Total mass of bullets in 1 second mb = (8 x 20)/1000 = 0.16 kg

F = [(0.16 x 250) - 0]/1 = 40 N
QUESTION: 10

The momentum of an isolated system remains conserved provided:

Solution: If there is no force acting on the system, then there will be no change in the moment of the system, Hence the total momentum will remain conserved
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