Impulse | Impulse-Momentum theorem
Impulse | Impulse-Momentum theorem :-
The change in momentum of an object can be made in two ways-
- By applying a smaller magnitude of force for long time.
- Applying a large force on the object for a short time.
Thus we can say that the change in momentum of an object is proportional to the product of force applied and the time duration.
Sometimes a large force acts for a very short duration and produces a finite change in momentum of the body. For example, hitting a cricket ball by a bat. Here, the ball is reflected back and the force on the ball acts for a very short time when the two are in contact. Practically, it is not possible to measure either the magnitude of force or the time for which it acts. In such a case, the total effect of force is measured by the impulse of the force.
Impulse: – When a force of greater magnitude is applied on an object for a short time (for example, hitting the ball with the bat, hitting the nail with the hammer), the product of the force and time difference is called the impulse of the force.
# This is a vector quantity and it is expressed by the symbol I.
# The direction of impulse is in the direction of force.
Unit of impulse: –
Formula of impulse: –
If a force F is acting on an object for a short time dt, then the impulse of this force
dI = F dt
Total impulse I = ∫Fdt = Area of the graph between force and time
From Newton’s Second Law of Motion… ..
F = dP/dt
∴ Fdt = dP
⇒ ∫Fdt = ∫dP
⇒ I = P2 – P1
This implies that the Impulse of a force is equal to the change in momentum of the body due to that force. This is the impulse-momentum theorem.
Some applications of the concept of impulse
- While catching a cricket ball, the player lowers his hands.
- Vehicles like cars, trucks, bogies of trains etc. are provided with a spring system (known as shockers) to avoid severe jerks.
- China- clay wares and glass wares are wrapped in paper or straw pieces.
- An athlete is advised to come to stop slowly after finishing a fast race.
- Person falling on cemented floor is likely to receive more injuries than one falling on a heap of sand.
Example 1. A 1 kg ball drops vertically into the floor with a speed of 25 ms-1 and rebounces with a speed of 10 ms-1. What is the impulse acting on the wall ?
(a) 15 kg m/s (b) 35 kg m/s (c) 45 kg m/s (d) 60 kg m/s
Solution :- As I = P2 – P1
Let us assume vertically downward direction as negative and vertically upward direction as positive.
P1 = (-) 1kg × 25 ms-1 = -25 kg m/s
P2 = (+) 1kg × 10 ms-1 = +10 kg m/s
∴ P2 – P1 = +10 kg m/s – [-25 kg m/s] = + 35 kg m/s
Hence correct option is (b).
Example 2. A player catches a cricket ball of mass 150 g moving at 20m/s. If the catching process is completed in 0.1s, the force exerted by the ball on the hands is
(a) 3000 N (b) 300 N (c) 30 N (d) 30000 N
Solution :- Again I = P2 – P1
P2 = Zero (as the ball stops after the catch)
P1 = (150/1000)kg × 20 ms-1 = 3 kg m/s
Now force exerted on hands, F = ΔP/Δt = (3 kg m/s) / (0.1s) = 30 kg m/s2 = 30 N
Hence correct option is (c).
Example 3. The area of the force-time graph gives
(a) velocity (b) acceleration (c) displacement (d) change in momentum
Solution :- We know that
Total impulse I = ∫Fdt = Area of the graph between force and time = Change in momentum
Hence correct option is (d).
Example 4. A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block will be :
(a) 2.5 m/s2 (b) 5 m/s2 (c) 10 m/s2 (d) 5 m/s2
Solution :- The final velocity of mass of water after hitting the metal block is zero so
average change in momentum/sec = initial momentum = mass × initial velocity of water
Δp = =1kg×5m/s=5kgm/s
As this change in momentum occurs during 1 sec, we may say that the force exerted on the 2kg metal block is
F=change in momentum/time =(5kgm/s)/1s = 5N
Hence by Newton’s Second law, initial acceleration of the block,
a = F/m = 5N/2kg = 2.5 m/s2
Hence correct option is (a).
For variable mass system,
Here, rate of change of mass
And v = 5 m/s
So F = 5 × 1 = 5 N
And acceleration, a = a = F/m = 5N/2kg = 2.5 m/s2
Example 5. A constant force acts on a body of mass 50 g at rest for 2 seconds. If the body moves through 27 m during that time, impulse of the force is
(a) 1.35 kg m/s (b) 13.5 Ns (c) 135 Ns (d) 2.7 kg m/s
Solution :- Impulse = change in momentum(ΔP) = P2 – P1
Using S = ut + (1/2)at2
27 = 0×2 + (1/2)a×(2)2
⇒ a = (27/2) m/s2
Now F = ma = (50/1000) × (27/2) = 0.675 N
Impulse = Ft = 0.675 × 2 = 1.35 kg m/s
Hence correct option is (a).
Example 6. A ball strikes a bat with a force 400N. If both are in contact for 0.01 sec. the impulse of the force
(a) 400 Ns (b) 4 Ns (c) 40000 Ns (d) 0.004 Ns
Solution :- Impulse = Ft = 400 × 0.01 = 4 Ns
Example 7. A body of mass 40 kg, moves with a uniform velocity under the action of a force 50 N on a surface. If a force of 70 N now acts on the same body in the same direction as that of 50 N, moving on the same surface, the acceleration of the body is
(a) 1.75m/s2 (b) 1.5 m/s2 (c) 1.0 m/s2 (d) 0.5 m/s2
Example 8. A body moving with constant velocity is brought to rest in 0.25 sec by applying a retarding force of 100N. The initial momentum of the body
(a) 25 kg m/s (b) 50 kg m/s (c) 100 kg m/s (d) 125 kg m/s
Example 9. A 2 kg plate is kept suspended in air by allowing 10 marbles hitting per second with a speed v from the downward direction. If the mass of each marble is 20 g then determine v.
(a) 20 ms-1 (b) 100 ms-1 (c) 40 ms-1 (d)50 ms-1
Example 10. Two stones of masses m1 and m2 are let fall from heights 2h and h, their momenta on reaching the ground are in the ratio
(a) 1 : 1 (b) 𝑚1: √2𝑚2 (c) √2𝑚1: 𝑚2 (d) 𝑚1: 𝑚2
Example 11. A body of mass 100 kg is moving with a velocity 1 m/s. The frictional force offered by the surface is 5 kg wt. If the body is pushed by a force of 50 N for one minute, velocity of the body after one minute is
(a) 0.16 m/s (b)16 m/s (c) 1.6m/s (d) 3.2 m/s
Example 12. A body of mass 1.5 kg falls vertically downwards with an acceleration of 29.4 m/s2. The force acting on the body in addition to force of gravity is
(a) 9.8 N (b) 19.6 N (c) 29.4 N (d) 49 N
Example 13. A body of mass 5 kg started from rest with an acceleration of 4 ms-2. Its momentum after 5s is
(a) 20 kgms-1 (b) 100 kgms-1 (c) 4 kgms-1 (d) 25 kgms-1
Example 14. A 6 kg balls strikes a vertical wall with a velocity 34 ms-1 and rebounds with a velocity of 26 ms-1. The impulse is
(a) 60 Ns (b) 180 Ns (c) 48 Ns (d) 360 Ns
Example 15. A block of a mass 4 kg is sliding on a smooth inclined plane of inclination 30°. Its momentum after 2 sec is
(a) 0 kgms-1 (b) 39.2 kgms-1 (c) 19.6 kgms-1 (d) 9.8 kgms-1
Example 16. Two bodies of mass 5 kg and 20 kg at rest are acted upon by the same force. The ratio of the times for which the force must act to produce the same impulse is
(a) 1 : 1 (b) 1 : 4 (c) 4 : 1 (d) 1 : 16