Physics numerical :- Force and Momentum
Q.1 . A force of 12 N on an object produces an acceleration of 4 m/s. How much force is needed to produce an acceleration of 10 m/s?
A) 30N
B) 12N
C) 10N
D) 20N
ANSWER= A ) 30N
Explain:- F = ma
m = F/a
m = 12/4
m =3 kg
Therefore, force for an acceleration of 10 m/s² = ma
= 3×10
= 30N ans.
Q.2. The momentum of an object is 20 kgm/s. How much force is needed to stop it in 4 seconds :-
A) 4N
B) 12N
C) 10N
D) 5N
ANSWER= D) 5N
Explain:- From the formula of impulse,
force × time = change in momentum
Now, initial momentum = 20 kgm/s
Final momentum = 0 kgm/s (momentum also becomes zero when velocity is zero)
Change in momentum = 20 - 0
= 20 kgm/s
Therefore, F× 4=20
F = 20/4
= 5N ans.
Q.3. An 80 g bullet is fired from a 20 kg gun with a velocity of 160 m/s. Find the velocity of the gun backwards.
A) 0.54 M/S
B) 0.64 M/S
C) 0.70 M/S
D) 0.39 M/S
ANSWER= B) 0.64
Explain:- Momentum of gun = Momentum of bullet
20 kg × v = 0.08 × 160
v = 12.8 20
v = 0.64 m/s ans.
Q.4. A force of 50 N is applied on an object for 8 seconds. What will be the change in momentum :-
A) 400 Nm
B) 500 Nm
C) 300 Nm
D) 250 Nm
ANSWER= A) 400 Nm
Explain:- Change in momentum = Force × Time
= 50 × 8
= 400 Nm ans.
5. A force of 200 dynes is applied on a 10 g object for 5 seconds. If the object was at rest, what would be the final velocity?
A) 80 cm/s
B) 50 cm/s
C) 100 cm/s
D) 250 cm/s
ANSWER= C) 100 m/S
Explain:- From the formula F = ma ,
200 = 10 × a
a = 200÷10
a = 20 cm/s²
Now, a = 20 cm/s² i.e., there is an increase of 20 cm/s in 1 second.
Total velocity increase in 5 seconds = 5 × 20
=100 cm/s
6. An object of mass 10 kg is moving with a velocity of 100 m/s and a force is applied in the opposite direction. If the object comes to rest in 10 seconds, find the acceleration produced by the force.
A) 8 m/s²
B) 5 m/s²
C) 10 m/s²
D) 25 m/s²
ANSWER= C) 10 m/s²
Explain:- At what rate should the 100 m/s velocity be eliminated so that it becomes 0 m/s in 10 seconds.
ans = 100/10
= 10
i.e. a = 10m/s²
7. A constant force is applied on a 2 kg moving object due to which the object travels a distance of 3 m in the first second and 1 m in the second second. How much force is applied on the object :-
A) 4N
B) 10N
C) 2N
D) 8N
ANSWER= A) 4N
Explain:- Velocity of object in first second = 3 m/s
Velocity of object in second second = 1 m/s
(from velocity = distance÷time formula)
Decrease in velocity in one second = (3 - 1) m/s
= 2 m/s
There is an increase or decrease in velocity in one second.
Hence acceleration = 2 m/s²
force = mass × acceleration
= 2 × 2
=4 N ans.
8. A force of 5 N is applied to an object of mass 9.8 N. Find the acceleration produced on the object.
A) 4 m/s²
B) 10 m/s²
C) 49 m/s²
D) 5 m/s²
ANSWER= D) 5 m/s²
Explain:- weight = mass × g
9.8 = mass × 9.8
Therefore, mass = 9.8/9.8
= 1kg
From the formula F =ma,
a = F/m
a = 5÷1
= 5 m/s²
9. A 1000 kg vehicle is moving with a velocity of 10 m/s. Find the amount of force which can increase the velocity of the vehicle by 25 m/s in 5 seconds.
A) 4000N
B) 1000N
C) 4900
D) 3000N
ANSWER= D)3000N
Explain:- Total increase in velocity in 5 seconds = (25-10)m/s
= 15 m/s
Total increase in velocity in 1 second = 15÷5
= 3 m/s
The increase in velocity in 1 second is called acceleration.
Hence acceleration = 3 m/s²
force = ma
= 1000×3
= 3000N ans.
10. A bullet of mass m moving with velocity v collides with a piece of wood of mass M and enters inside the wood. Find the velocity produced in the wood.
A)( M+m/m)v
B) (m +M)v
C) (m/M+m) ×v
D) v(m/M)
ANSWER= C) (m/M+m) ×v
Explanation:- Before hitting, the momentum of the bullet = mv
Mass of wood after collision = (m+M)
Let velocity of (bullet + wood) after collision = y
Hence momentum of (bullet + wood) = (m+M)y
By the law of conservation of momentum,
(m+M)y = mv
y = (mv)÷(m+M)
11. An object of 5 kg is moving with an acceleration of 0.30 m/s². Find the amount of force acting on it.
A) 2.3N
B) 1.5N
C) 3.5N
D) 2N
ANSWER= B) 1.5N
Explain:- F = ma
= 5 × 0.30
= 1.5 N ans.
12. A force of 10N is applied on a 2 kg object for 10 seconds. Find the velocity of the object after 10 seconds.
A) 27 M/S
B) 40 M/S and
C) 30 M/S
D) 50 m/s
ANSWER= D) 50 m/s
Explain:- F = ma
10 = 2×a
a = 5m/s²
The increase in velocity in 1 second is the acceleration.
Therefore increase in velocity in 1 seconds = 5 m/s
Total increase in velocity in 10 seconds = 5 × 10 m/s
= 50 m/s
13. A car of mass 1120 kg is moving with a speed of 80 km/h. Brakes have been applied on it and the car comes to a halt after covering a distance of 70 km. How much force is applied to stop the car.
A) 5N
B) 2.5N
C) 3.95N
D) 4N
ANSWER= C) 3.5N
Explain:- • Work done in stopping the car = Kinetic energy of the car
• Work = Force × Distance
Velocity = 80 km/h
= (80 ×1000)÷ 3600 m/s
= 200÷9 m/s
Work = ×m ×v²
= ×1120 ×(200/9)²
= 560 ×( 40000÷81) j
work = force × distance
560 × (40000÷81) = force × 70000
Force = {( 560 × 40000 )÷ (70000 × 81)
= 320÷ 81
= 3.95N
14. A bullet of 0.015 kg moving with a velocity of 1000 m/s strikes a wood and penetrates 0.1 m. How much force is applied to stop the bullet?
A) 4.6 x 10 N
B) 8.2 x10 N
C) 6.5 X 10N
D) 7.5 × 10⁴N
ANSWER= D) 7.5 × 10⁴N
Explain:- Kinetic energy = × 0.015 ×(1000)² j
= 7.5 × 1000 j
kinetic energy = work = force × distance
7.5 × 1000 = force × 0.1
Force = 7.5 × 1000 × 10
= 7.5 × 10⁴ N
15. A car is moving with a velocity of 108 km/h. After applying the brakes it stops in 4 s. Calculate the force exerted by the brakes on the car. The total mass of the car is 1000kg.
A) 750 N
B) 700 N
C) 650 N
D) 475 N
ANSWER= A)750N
Explain:- v = 108 km/h
={(108×1000)÷3600} m/s
= 30 m/s
In 4s, 30 m/s velocity is lost.
So decrease in velocity in 1 second = 30÷4
= 7.50 m/s
There is a decrease or increase in velocity in 1 second.
Therefore a = 7.50 m/s²
F = ma
= 1000 × 7.5
= 750 N ans.
16. A 20 g bullet is fired from a 2 kg gun with a velocity of 150 m/s. Find the recoil velocity of the gun.
A) 0.50 m/s
B) 1.5 m/s
C) 3 m/s
D) 1.75 m/s
ANSWER= B)1.5 m/s
Explain:- Mass of the bullet = 20 g
= 0.02 kg
By the law of conservation of momentum, momentum of the gun = momentum of the bullet
2 × v = 0.02 × 150
v = 1.5 m/s
17. The velocity of a 2 kg object is 10 m/s and that of a 3 kg object is 8 m/s. After the collision of both the objects the velocity of 2kg object is 12 m/s, then what will be the velocity of 3kg object?
A) 7 m/s
B) 1.5 m/s
C) 7.5 m/s
D) 6.7 m/s
ANSWER= D) 6.7 m/s
Explain:- Total momentum of both the objects before colliding = (2×10)+(3×8)
= 20+24
=44
Let velocity of 3kg object after collision =v
Total momentum of both the objects after collision = (2×12)+(3×v)
=24+3v
By the law of conservation of momentum,
Total momentum before the collision = Total momentum after the collision
44= 24+3v
3v= 44-24
3v =20
v= 20÷3
= 6.7 m/s
18. A rocket of mass 1000 kg expels gases at the rate of 4 kg/s. If the velocity of the gases is 3000 m/s, calculate the force exerted by the gas on the rocket.
A) 12000N
B) 10000N
C) 60000N
D) 70000N
ANSWER= D) 12000N
Explain: - 4kg of gases are being changed from 0m/s to 3000 m/s velocity per second.
Change in momentum = (4×3000 - 4×0)kgm/s
= 12000 kgm/s
force = change in momentum time
Force = (12000÷ 1 )N
= 12000N
Force per second applied on the rocket = force produced in 4kg of gases
19. A 10 gm bullet was fired from a gun of mass 1 kg. If the recoil velocity of the gun is 5 m/s then find the velocity of the bullet:-
A) 350 m/s
B) 550 m/s
C) 500 m/s
D) 400 m/s
ANSWER= C) 500 m/s
Explain:- Before firing the bullet,
Gun velocity = 0 m/s
Bullet velocity = 0 m/s
Hence total momentum (gun + bullet) = 0
After the shot,
Total momentum = momentum of gun + momentum of bullet
= 1 × 5 + 0.01× v
= 5 + 0.01v
By the law of conservation of momentum,
5 +0.01v = 0
0.01v = -5
v = -5× 100
= -500m/s
20. A rocket of 600 kg is set for vertical launch. If the velocity of the gases released by the rocket is 1000 m/s then how many kg/s of gases are required to be burnt to balance the weight of the rocket:-
A) 6kg
B) 4kg
C) 1kg
D) 3kg
ANSWER= A) 6kg
Explain:- On taking g = 10 m/s²,
Weight of rocket = mg
= 600×10
= 6000N
Now to balance the 6000 N load, the force of the gases released per second should be 6000N . Let the gas released on burning m kg of gas hold a force of 6000N.
From Newton's second law,
F =m(v -u)÷t
6000 = m( 1000 -0)÷1
m = 6000÷1000
=6kg
(Note: Here 6kg of gases are being changed from 0 m/s to 1000 m/s in 1 second.)
21. A ship of mass 3 × 10⁷ kg is pulled by a force of 5 × 10⁴ N for 3 m. What will be the speed of the ship if the friction of water is taken to be zero? :-
A) 0.75 m/s
B) 4 m/s
C) 1 m/s
D) 0.1 m/s
ANSWER= D) 0.1 m/s
Explanation:- Work = Kinetic energy (both are form of energy)
Work = Force × Distance
= 5 ×10⁴ ×3
= 15× 10⁴
Hence kinetic energy = 15 × 10⁴
(½)mv²= 15 ×10⁴
(½)×3×10⁷ ×v² = 15× 10⁴
v² = (10÷1000)
v =√(1÷100)
= 1/10
= 0.1 m/s
22. A toy rocket releases gases at the rate of 0.05 kg per second. If the velocity of the gases is 400m/s then find the force on the rocket:-
A) 10 N
B) 15 N
C) 18 N
D) 20 N
ANSWER= D) 20N
Explain:- time t= 1 second m = 0.05 kg u= 0 m/s v = 400 m/s
From Newton's second law,
F = m(v-u)÷t
= 0.05(400-0)÷1
= 20.00 N
(Note :- The velocity of 0.05 kg of gases per second is being changed from 0m/s to 400 m/s.)
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