Two balls of equal mass are thrown at a wall. One is made of rubber and bounces back, while the other is made of clay and sticks to the wall. Which ball experiences a greater impulse? Explain.

The rubber ball experiences a greater impulse. Impulse is the change in momentum. The clay ball's momentum changes from its initial value to zero. The rubber ball's momentum changes from its initial value to zero, and then changes again as it bounces back in the opposite direction. This greater change in momentum means a greater impulse.

Two identical discs are rotating with different angular velocities. Which disc experiences a greater angular impulse if they are both brought to rest by the same constant braking torque?

The disc with the higher initial angular velocity experiences a greater angular impulse because it takes longer to stop, even though both disks experience the same torque and angular acceleration. Angular impulse is the torque multiplied by the time interval, and the disc with the higher initial angular velocity has a longer stopping time.

How does the concept of angular impulse relate to the design of a baseball bat?

A baseball bat is designed to deliver a large angular impulse to the ball. A heavier bat, swung with the same angular velocity, has a larger initial angular momentum. If the bat can transfer a large fraction of this angular momentum to the ball during the collision, the ball will leave with a greater linear velocity. The impact force between the bat and ball, multiplied by the contact time, is the angular impulse delivered to the ball (and an equal and opposite angular impulse is delivered to the bat). A larger angular impulse on the ball translates to a greater change in the ball's momentum (and therefore a greater velocity).

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Impulse

Impulse is like a sudden push or pull that can change how an object is moving, whether it's moving in a straight line or spinning around. It helps us understand what happens when things bump into each other, like in sports or car crashes. Impulse is important for designing machines and understanding how they work. Since it has both magnitude and direction, we need to consider both to understand its effects.

1.0Definition of Impulse

When a large force is applied on a body for a very short interval of time, then the product of force and time interval is known as impulse.

$d I = F d t = d P [[\frac{d p}{d t}] = F]$

2.0 Units And Examples of Impulse

1.Airbags in Cars:Airbags reduce impact force during a collision by increasing the collision time, which reduces the impulse needed to stop the passenger, making it safer.

2.Sports: In sports like cricket or baseball, a bat applies force to the ball over a period of time, giving the ball a large change in momentum, making it travel at high speeds.

3.Buffers are provided in train bogies to avoid jerks.

4.A person jumping on a hard cement floor receives more injury than a person jumping on a muddy or sandy road.

Unit of impulse = N-s or kg-m/s.

Dimension of impulse $[F] [t] = [m] [a] [t] = [M L T^{- 2} T] = [M L T^{- 1}]$

3.0Analysis Of Impulse And Average Force

Case-1: If this force is acting from time $t_{1} and t_{2}$ , then integrating the above equation, we get

$I = \int_{t_{1}}^{t_{2}} F d t = F \int_{p_{1}}^{p_{2}} d t$

Impulse = Change in momentum

Case-2:If constant or average force acts on a body, then :-

$I = \int_{t_{1}}^{t_{2}} F d t = \int_{p_{1}}^{p_{2}} d p \Rightarrow I = F \int_{p_{1}}^{p_{2}} d p$

$I = F_{avg} \int_{t_{1}}^{t_{2}} d t = F_{avg} (t_{2} - t_{1}) = (p_{2} - p_{1}) \Rightarrow I = F_{avg} Δ t = Δ p$

Case-3: Impulse from force time graph :

The area under the force time graph represents impulse or change in momentum.

$I = Δ p = \int F d t = Area Under F - t Graph$

Average Force: Average force acting on a body in a given time interval can be calculated by,

$\frac{d p}{d t} = F = F_{avg} = \frac{Δ p}{Δ t}$

$F_{avg} = \frac{Δ p}{Δ t}$

Therefore, for a certain momentum change if the time interval is increased, then the average force exerted on the body will decrease.

4.0Impulse Momentum Theorem

This theorem states that the impulse exerted on an object is equivalent to the change in its momentum.This theorem can be derived from Newton's Second Law of Motion. Since the rate of change of momentum is instantaneous, the impulse is effectively equal to the change in momentum.

$F = \frac{d p}{d t} ∴ \frac{d p}{d t} \Rightarrow Rate of change of momentum with respect to time$

$I = \int F d t$ (Impulse)

$\int F d t = \int d p$

$\int_{0}^{t} F d t = \int_{p_{1}}^{p_{2}} d p$

$F (t - 0) = (p_{2} - p_{1})$

$I = \int F d t = (p_{2} - p_{1})$

Note:Impulse applied to an object will be equal to the change in its momentum.

Examples:A body of mass 3 kg has an initial speed $4 m s^{- 1}$ . A force acts on it for some time in the direction of motion. The force-time graph is

shown in figure. The final speed of the body is :

Solution:

p=Area under F-t graph

$m (v - u) = 4 + 8 + 1.625 + 5 \Rightarrow 3 (v - 4) = 18.625 \Rightarrow v = 10.208 m / s$

5.0Application of Impulse

1.Liquid Jet Striking a Vertical Wall Normally

Consider a liquid jet with an area A striking a wall at a velocity v. After impacting the wall, the liquid moves parallel to the surface of the wall.
Consider a small element of liquid with mass m.The change in momentum after it strikes the wall is given by

$[Δ p]_{x}^{jet} = 0 - Δ m v = - Δ m v$

$[Δ p]_{x}^{jet} + [Δ p]_{x}^{wall} = 0$

Transfer of momentum to the $[Δ p_{x}]_{wall} = Δ m v$

Force applied by the jet on the wall, $F = \frac{[ Δ p _{x} ] _{wall}}{Δ t} = \frac{Δ m v}{Δ t}$

∵ $Δ m = ρ a v Δ t$

$F = \frac{[ Δ p ] _{x}^{wall}}{Δ t} = \frac{Δ m v}{Δ t} = (\frac{Δ m}{Δ t}) v = (ρ \frac{Δ V}{Δ t}) v = ρ (\frac{Δ V}{Δ t}) v$

Rate of flow of liquid is $\frac{Δ V}{Δ t} = a v$

$F = ρ a v^{2}$

2.Liquid Jet striking the wall at the same angle

The Jet strikes the wall at an angle $θ$ to the normal and rebounds with the same angle.

Transfer of Momentum to the Wall,

$[Δ p]_{x}^{wall} = 2Δ m v cos θ, [Δ p]_{y}^{wall} = 0$

Force applied by the jet on the wall, $F = \frac{[ Δ p ] _{x}^{wall}}{Δ t} = \frac{2Δ m v c o s θ}{Δ t} = (\frac{Δ m}{Δ t}) 2 v cos θ$

$F = ρ (\frac{Δ V}{Δ t}) 2 v cos θ = ρ (a v) 2 v cos θ = 2 ρ a v^{2} cos θ$

6.0Angular Impulse

Angular momentum is a mathematical quantity that describes the rotational effect of a force (torque) acting on a body over a given period of time.
The change in angular momentum of a system resulting from the action of a torque over an infinitesimal time interval is known as the angular impulse imparted to the system by the torque.

$J = \int τ d t$

$J = τ_{avg} \int d t$

$J = L_{f} - L_{i}$

$J = I (ω_{f} - ω_{i}) = I (ω_{f} - ω_{i})$

As linear impulse changes the linear momentum,angular impulse changes the angular momentum.

$J = \int τ d t$ =angular impulse is equal to the changes in angular momentum.

Example:A solid ball of mass ’m’ and radius ‘R’ is kept over rough ground.A time varying force F=2t is acting at the topmost point as shown in the figure.Find angular momentum of the sphere about the bottommost point as a function of time ‘t’.

Solution:

Suppose ‘f’ is the frictional force acting on the sphere in the forward direction, as illustrated in the figure.

Taking the torque about the bottommost point O, the torque due to friction is zero, as the frictional force acts through point O.

= $τ = F \times r_{⊥} = (2 t) (2 R) = 4 Rt$

Angular Impulse $\int_{0}^{t} t d t = \int_{0}^{t} 4 Rt d t = 2 R t^{2}$

7.0Solved Examples

1 .A football of mass 0.8 kg coming towards the player with a velocity of 5 m/s. The player hit straight back the football with velocity 10 m/s. If the contact time is 0.02 second then Find the average force involved?

Solution:

$F_{avg} = \frac{Δ p}{Δ t} = \frac{0.8 ( 10 - ( - 5 ))}{0.02} = 600 N$

2. A Mallet of mass 1 kg moving with a speed of 6 m/s blows a wall and comes to downtime in 0.1 s. Calculate

(a) Impulse of the force

(b) Average retarding force that stops the hammer

(c) Average retardation of the hammer

Solution:

(a) Impulse= $F \times Δ t = m (v - u) = 1 (0 - 6) = - 6 N \cdot s$

(b) Average retarding force, $F = \frac{Impulse}{Time} = \frac{6}{0.1} = 60 N$

(c) Average retardation, $a = \frac{F}{m} = \frac{60}{1} = 60 m / s^{2}$

Impulse

Impulse is like a sudden push or pull that can change how an object is moving, whether it's moving in a straight line or spinning around. It helps us understand what happens when things bump into each other, like in sports or car crashes. Impulse is important for designing machines and understanding how they work. Since it has both magnitude and direction, we need to consider both to understand its effects.

1.0Definition of Impulse

When a large force is applied on a body for a very short interval of time, then the product of force and time interval is known as impulse.

$d I = F d t = d P [[\frac{d p}{d t}] = F]$

2.0 Units And Examples of Impulse

1.Airbags in Cars:Airbags reduce impact force during a collision by increasing the collision time, which reduces the impulse needed to stop the passenger, making it safer.

2.Sports: In sports like cricket or baseball, a bat applies force to the ball over a period of time, giving the ball a large change in momentum, making it travel at high speeds.

3.Buffers are provided in train bogies to avoid jerks.

4.A person jumping on a hard cement floor receives more injury than a person jumping on a muddy or sandy road.

Unit of impulse = N-s or kg-m/s.

Dimension of impulse $[F] [t] = [m] [a] [t] = [M L T^{- 2} T] = [M L T^{- 1}]$

3.0Analysis Of Impulse And Average Force

Case-1: If this force is acting from time $t_{1} and t_{2}$ , then integrating the above equation, we get

$I = \int_{t_{1}}^{t_{2}} F d t = F \int_{p_{1}}^{p_{2}} d t$

Impulse = Change in momentum

Case-2:If constant or average force acts on a body, then :-

$I = \int_{t_{1}}^{t_{2}} F d t = \int_{p_{1}}^{p_{2}} d p \Rightarrow I = F \int_{p_{1}}^{p_{2}} d p$

$I = F_{avg} \int_{t_{1}}^{t_{2}} d t = F_{avg} (t_{2} - t_{1}) = (p_{2} - p_{1}) \Rightarrow I = F_{avg} Δ t = Δ p$

Case-3: Impulse from force time graph :

The area under the force time graph represents impulse or change in momentum.

$I = Δ p = \int F d t = Area Under F - t Graph$

Average Force: Average force acting on a body in a given time interval can be calculated by,

$\frac{d p}{d t} = F = F_{avg} = \frac{Δ p}{Δ t}$

$F_{avg} = \frac{Δ p}{Δ t}$

Therefore, for a certain momentum change if the time interval is increased, then the average force exerted on the body will decrease.

4.0Impulse Momentum Theorem

This theorem states that the impulse exerted on an object is equivalent to the change in its momentum.This theorem can be derived from Newton's Second Law of Motion. Since the rate of change of momentum is instantaneous, the impulse is effectively equal to the change in momentum.

$F = \frac{d p}{d t} ∴ \frac{d p}{d t} \Rightarrow Rate of change of momentum with respect to time$

$I = \int F d t$ (Impulse)

$\int F d t = \int d p$

$\int_{0}^{t} F d t = \int_{p_{1}}^{p_{2}} d p$

$F (t - 0) = (p_{2} - p_{1})$

$I = \int F d t = (p_{2} - p_{1})$

Note:Impulse applied to an object will be equal to the change in its momentum.

Examples:A body of mass 3 kg has an initial speed $4 m s^{- 1}$ . A force acts on it for some time in the direction of motion. The force-time graph is

shown in figure. The final speed of the body is :

Solution:

p=Area under F-t graph

$m (v - u) = 4 + 8 + 1.625 + 5 \Rightarrow 3 (v - 4) = 18.625 \Rightarrow v = 10.208 m / s$

5.0Application of Impulse

1.Liquid Jet Striking a Vertical Wall Normally

Consider a liquid jet with an area A striking a wall at a velocity v. After impacting the wall, the liquid moves parallel to the surface of the wall.
Consider a small element of liquid with mass m.The change in momentum after it strikes the wall is given by

$[Δ p]_{x}^{jet} = 0 - Δ m v = - Δ m v$

$[Δ p]_{x}^{jet} + [Δ p]_{x}^{wall} = 0$

Transfer of momentum to the $[Δ p_{x}]_{wall} = Δ m v$

Force applied by the jet on the wall, $F = \frac{[ Δ p _{x} ] _{wall}}{Δ t} = \frac{Δ m v}{Δ t}$

∵ $Δ m = ρ a v Δ t$

$F = \frac{[ Δ p ] _{x}^{wall}}{Δ t} = \frac{Δ m v}{Δ t} = (\frac{Δ m}{Δ t}) v = (ρ \frac{Δ V}{Δ t}) v = ρ (\frac{Δ V}{Δ t}) v$

Rate of flow of liquid is $\frac{Δ V}{Δ t} = a v$

$F = ρ a v^{2}$

2.Liquid Jet striking the wall at the same angle

The Jet strikes the wall at an angle $θ$ to the normal and rebounds with the same angle.

Transfer of Momentum to the Wall,

$[Δ p]_{x}^{wall} = 2Δ m v cos θ, [Δ p]_{y}^{wall} = 0$

Force applied by the jet on the wall, $F = \frac{[ Δ p ] _{x}^{wall}}{Δ t} = \frac{2Δ m v c o s θ}{Δ t} = (\frac{Δ m}{Δ t}) 2 v cos θ$

$F = ρ (\frac{Δ V}{Δ t}) 2 v cos θ = ρ (a v) 2 v cos θ = 2 ρ a v^{2} cos θ$

6.0Angular Impulse

Angular momentum is a mathematical quantity that describes the rotational effect of a force (torque) acting on a body over a given period of time.
The change in angular momentum of a system resulting from the action of a torque over an infinitesimal time interval is known as the angular impulse imparted to the system by the torque.

$J = \int τ d t$

$J = τ_{avg} \int d t$

$J = L_{f} - L_{i}$

$J = I (ω_{f} - ω_{i}) = I (ω_{f} - ω_{i})$

As linear impulse changes the linear momentum,angular impulse changes the angular momentum.

$J = \int τ d t$ =angular impulse is equal to the changes in angular momentum.

Example:A solid ball of mass ’m’ and radius ‘R’ is kept over rough ground.A time varying force F=2t is acting at the topmost point as shown in the figure.Find angular momentum of the sphere about the bottommost point as a function of time ‘t’.

Solution:

Suppose ‘f’ is the frictional force acting on the sphere in the forward direction, as illustrated in the figure.

Taking the torque about the bottommost point O, the torque due to friction is zero, as the frictional force acts through point O.

= $τ = F \times r_{⊥} = (2 t) (2 R) = 4 Rt$

Angular Impulse $\int_{0}^{t} t d t = \int_{0}^{t} 4 Rt d t = 2 R t^{2}$

7.0Solved Examples

1 .A football of mass 0.8 kg coming towards the player with a velocity of 5 m/s. The player hit straight back the football with velocity 10 m/s. If the contact time is 0.02 second then Find the average force involved?

Solution:

$F_{avg} = \frac{Δ p}{Δ t} = \frac{0.8 ( 10 - ( - 5 ))}{0.02} = 600 N$

2. A Mallet of mass 1 kg moving with a speed of 6 m/s blows a wall and comes to downtime in 0.1 s. Calculate

(a) Impulse of the force

(b) Average retarding force that stops the hammer

(c) Average retardation of the hammer

Solution:

(a) Impulse= $F \times Δ t = m (v - u) = 1 (0 - 6) = - 6 N \cdot s$

(b) Average retarding force, $F = \frac{Impulse}{Time} = \frac{6}{0.1} = 60 N$

(c) Average retardation, $a = \frac{F}{m} = \frac{60}{1} = 60 m / s^{2}$

Frequently Asked Questions

Two balls of equal mass are thrown at a wall. One is made of rubber and bounces back, while the other is made of clay and sticks to the wall. Which ball experiences a greater impulse? Explain.

Two identical discs are rotating with different angular velocities. Which disc experiences a greater angular impulse if they are both brought to rest by the same constant braking torque?

How does the concept of angular impulse relate to the design of a baseball bat?

Join ALLEN!

Impulse

1.0Definition of Impulse

2.0 Units And Examples of Impulse

3.0Analysis Of Impulse And Average Force

4.0Impulse Momentum Theorem

5.0Application of Impulse

6.0Angular Impulse

7.0Solved Examples

Table of Contents

Frequently Asked Questions

Two balls of equal mass are thrown at a wall. One is made of rubber and bounces back, while the other is made of clay and sticks to the wall. Which ball experiences a greater impulse? Explain.

Two identical discs are rotating with different angular velocities. Which disc experiences a greater angular impulse if they are both brought to rest by the same constant braking torque?

How does the concept of angular impulse relate to the design of a baseball bat?

Join ALLEN!

Impulse

1.0Definition of Impulse

2.0 Units And Examples of Impulse

3.0Analysis Of Impulse And Average Force

4.0Impulse Momentum Theorem

5.0Application of Impulse

6.0Angular Impulse

7.0Solved Examples

Table of Contents