Assume your file cabinet is sitting on a cart with wheels. It is sitting in
the middle of a room with a smooth floor. You want to move it against the
wall. You give it a push. It takes off, and before you know it, it slams into
the wall. It is hard to stop it, because it has **linear
momentum**.

Linear momentum is a measure of an object's translational
(as opposed to rotational) motion. The linear momentum **p** of an
object is defined as the product of the object's mass m times its velocity
**v**.

**p** = m**v.**

Linear momentum is a vector**.** Its direction is the direction of the velocity.

The Cartesian components of **p** are p_{x }= mv_{x}, p_{y }= mv_{y},
p_{z }= mv_{z}.

If an object's velocity is changing, its linear momentum is changing. For an object with constant mass we have

lim_{∆t-->0 }∆**p**/∆t = m
lim_{∆t-->0 }∆**v**/∆t = m d**v**/dt = m**a** =
**F.**

To change the linear momentum of an object of m, a net force has to act on
it, causing the object to accelerate. **F** = d**p**/dt
is a different way of stating
**Newton's second law**. The
rate at which an object's momentum changes is equal to the force acting on the
object. If a force **F** acts on an object for a time ∆t,
then the change in the object's momentum is ∆**p**
= **F**_{avg}∆t = ∫_{ti}^{tf
}
**F**dt. If ∆t
is a very short time interval, we say that the object receives an**
impulse ** **I **= ∆**p **=
**F**_{avg}∆t.

Note the difference:

Work: W = **F**_{avg}**·d** (scalar)

Impulse: **I **= ∆**p **= **F**_{avg}∆t (vector)

What is the magnitude of the momentum of a fly if it is traveling with speed
1 m/s and has a mass of 0.0001 kg?

Solution:

The momentum of the fly is (0.0001 kg)(1 m/s) = 0.0001 kgm/s.

A bullet is accelerated down the barrel of a gun by hot gases produced in the
combustion of gun powder. What is the average force
exerted on a 0.03 kg bullet to accelerate it to a speed of 600 m/s in a time of
2.00 ms (milliseconds)?

Solution

Let the positive x direction point down the barrel.

The change in momentum of the bullet is ∆p = p_{f} = p_{i} =
(0.03 kg)*(600 m/s) = 18 kgm/s.

The momentum changes in ∆t = 2*10^{-3} s.

The average force exerted is F_{avg} = ∆p/∆t = (18 kgm/s)/(2*10^{-3} s) = 9*10^{3}
N.

The force is positive, it points in the +x-direction.

An object that has a small mass and an object that has a large mass have the
same momentum **p**. Which object has the largest kinetic energy?

Solution

The magnitude of each object's momentum is p.

The kinetic energy of an object is K = ½mv^{2} = ½m^{2}v^{2}/m
= p^{2}/(2m).

The object with the larger mass has the smaller kinetic energy.

An object that has a small mass and an object that has a large mass have the
same kinetic energy. Which mass has the largest magnitude of momentum p?

Solution:

The kinetic energy of an object is K = ½mv^{2}.

v^{2} = 2K/m, m^{2}v^{2} = p^{2} =
2Km. p = (2Km)^{1/2}.

The object with the larger mass has the larger momentum p.

A car is stopped for a traffic signal. When the light turns green, the car
accelerates, increasing its speed from 0 to 5.2 m/s in 0.832s. What linear
impulse and average force does a 70 kg passenger in the car experience?

Solution:

Assume the car accelerates in the x-direction. The final momentum of the
passenger is p_{f }= (70 kg)(5.2 m/s) = 364 kgm/s in the
x-direction.. The initial momentum is zero.

∆p_{
}= p_{f} - p_{i
}= 364 kgm/s = F_{avg}∆t. **F**_{avg }= (364
kgm/s)/(0.832 s) = 437.5 N in the x-direction.

A 3 kg steel ball strikes a wall with a speed of
10.0 m/s at an angle of 60^{o} with the surface. It bounces off with
the same speed and angle. If the ball is in contact with the wall for 0.2 s, what is the average force exerted on the ball by the wall?

Solution:

Let **i** denote the positive x-direction and
**j**
the positive y-direction; **i** and **j** are unit vectors, they are
direction indicators.

The ball's initial momentum is**p**_{i
}= p_{xi}**i **+ p_{yi}**j **= (3 kg 10 m/s)sin60^{o
}**i**+ (3 kg 10 m/s)cos60^{o }**j**.

Its final momentum is

**p**_{f }= p_{xf}**i **+ p_{yf}**j **= -(3 kg 10 m/s)sin60^{o
}**i **+ (3 kg 10 m/s)cos60^{o }**j**.

∆**p **=
**p**_{f }-** p**_{i }= -2(30 kgm/s)sin60^{o}** i
**= -(51.96 kgm/s)** i**.

∆**p** =
**F**_{avg} ∆t.
**F**_{avg }= -(51.96 kgm/s)** i**/(0.2 s) = -259.8** i**.

Does a large force always produce a larger impulse on a body than a smaller force does? Explain!

Solution:

The impulse is the integral of force over time, **I
**= ∆**p **= **F**_{avg}∆t.

A small force acting over a long time can produce a larger
impulse than a large force acting over a short time.