Tag Archives: motion example problem

Impulse and Momentum – Physics Example Problem

Desktop Momentum Balls Toy

This model of a common desktop toy shows the forces acting on the raised ball. The principles of impulse and momentum show how the momentum is transferred to each ball and the process repeats.

Impulse and momentum are physical concepts that are easily seen from Newton’s Laws of Motion.

Start with this equation of motion for constant acceleration.

v = v0 + at

v = velocity
v0 = initial velocity
a = acceleration
t = time

If you rearrange the equation, you get

v – v0 = at

Newton’s second law deals with with force.

F = ma

F = force
m = mass
a = acceleration

solve this for a and get

a = F/m

Stick this into the velocity equation and get

v – v0 = (F/m)t

Multiply both sides by m

mv – mv0 = Ft

The left side of the equation deals with momentum (often denoted by a lower-case p) and the right side is impulse (often denoted by an upper-case letter J).

Mass times velocity is known as momentum and force applied over time is called impulse.

Impulse and Momentum Example Problem

Question: A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.

a) What is the initial momentum of the mass?
b) What is the final momentum of the mass?
c) What was the force acting on the mass?
d) What was the impulse acting on the mass?

Impulse and Momentum Example Problem

Part a) What is the initial momentum?

Momentum is mass times velocity. Since the mass is at rest, the initial velocity is 0 m/s.

momentum = m⋅v = (50 kg)⋅(0 m/s) = 0 kg⋅m/s

Part b) What is the final momentum?

After the force is finished acting on the mass, the velocity is 3 m/s.

momentum = m⋅v = (50 kg)⋅(3 m/s) = 150 kg⋅m/s

Part c) What was the force acting on the mass?

mv – mv0 = Ft

From parts a and b, we know mv0 = 0 kg⋅m/s and mv = 150 kg⋅m/s.

150 kg⋅m/s – 0 kg⋅m/s = Ft
150 kg⋅m/s = Ft

Since the force was in effect over 3 seconds, t = 3 s.

150 kg⋅m/s = F ⋅ 3s
F = (150 kg⋅m/s) / 3 s
F = 50 kg⋅m/s2

Unit Fact: kg⋅m/s2 can be denoted by the derived SI unit Newton (symbol N)

F = 50 N

Part d) What was the impulse acting on the mass?

The impulse is the force multiplied by the time passed. It is also equal to the change in momentum over the same time period.

Ft = 50 N ⋅ 3 s
Ft = 150 Ns or 150 kg⋅m/s

The impulse was 150 kg⋅m/s.

These problems are relatively simple as long as you keep your units straight. Impulse and momentum should have the same units: mass⋅velocity or force⋅time. Check your units when you check your answer.

Another possible way to cause errors is to confuse your vector directions. Velocity and Force are both vector quantities. In this example, the mass was pushed in the direction of the final velocity. If another force pushed in the opposite direction to slow down the mass, the force would have a negative value compared to the velocity vector.

If you found this helpful, check out other Physics Example Problems.

Inertia Example Problem – Two Connected Blocks

Complex systems can cause difficulty for students. When two different systems are connected together, there are some factors they share in common. Identifying these connections can make solving problems easier. This example problem has a complex system of two blocks connected by a massless string.

Example Problem:
Two blocks are connected by a massless string around a frictionless pulley. Block A is sliding across a frictionless surface and is pulled by the second block as Block B falls.
a) What is the acceleration of the system?
b) What is the tension in the string?

This illustration shows the arrangement of the blocks.

2 blocks connected

This system is coupled together by the massless string. As Block A moves to the right a distance Δd in some time t, Block B moves down Δd. That means the block’s velocities are the same.

vAΔdt = vB

The velocity directions can be adjusted by the coordinate system chosen for each system. Since the velocities are always the same, their accelerations are the same.

a = aA = aB

Since the string is massless, the tension is uniform throughout the system. The tension pulling Block A to the side is the same as the tension pulling Block B upwards.

Let’s find the forces of both systems.

forces on 2 blocksStart with Block A. Block A is being accelerated in the positive x-direction.

ΣFx = T = mAa
ΣFy = N – mAg

Since Block A is not moving in the vertical direction, the sum of the forces in that direction is equal to 0.

ΣFy = N – mAg = 0
N = mAg

Now find the forces on Block B. Block B is being accelerated down in the positive y’-direction and no forces are acting in the x’-direction.

ΣFy’ = mBa
ΣFy’ = mBg – T

Set these two equations equal to each other

mBa = MBg – T

Now we have two equations with two unknown variables. The simplest way to solve this is to solve one equation for one of the variables and then substitute that result into the second equation to solve for the other variable. Let’s solve the last equation for T.

mBa = MBg – T
mBa – mBg = -T
T = mBg – mBa

Plug this expression into the force equation involving the tension of the string from block A and solve for acceleration.

T = mAa
mBg – mBa = mAa

Add mBa to both sides

mBg = mAa + mBa

Factor out the acceleration

mBg = (mA + mB)a

Divide both sides by (mA + MB)

a = mBg/(mA+mB)

Now that we have the answer to Part a of the question. We can use this to find the tension. Plug in the solution into one of the equations containing the tension. Let’s use the easy one:

T = mAa
T = mAmBg/(mA+mB)

Note how the acceleration will always be less than g. Also notice the tension will always be less than the weight of Block B (mBg). One common mistake in this type of problem is to assume the tension in the string is equal to the weight of Block B. This would only be true if the Block B was in equilibrium. Since the block is accelerating, it is not in equilibrium.

Equations of Motion – Constant Acceleration Intercept Example Problem

Motion in a straight line with constant acceleration is the simplest form of accelerated motion. Since acceleration is constant, the velocity changes at the same rate as time progresses. Position will change as the square of the time progressing. There are three basic formulas that will help with most homework problems dealing with motion under constant acceleration.

(1) x = x0 + v0t + ½at2

(2) v = v0 + at

(3) v2 = v02 + 2a(x – x0)

x is the distance travelled
x0 is the initial starting point
v is the velocity
v0 is the initial velocity
a is the acceleration
t is the time

This worked example problem shows how to use the equations of motion to find the time it takes a constantly accelerating body to intercept another body moving at a constant velocity.

Example Problem:
A speeding motorist travelling at 120 km/hr passes a stopped police car. The police car immediately begins to chase the speeder, accelerating at a constant 2.5 m/s2.
(a) How long does it take for the police car to intercept the speeder?
(b) How far did the police car travel before catching up to the speeder?
(c) How fast was the police car travelling when it intercepts the speeder?

This illustration shows the conditions of the vehicles at the beginning of the problem and the time when the police car intercepts the speeder.

Constant Acceleration Intercept Setup

Top: Speeder moving at VS0 passes police car at rest at x0.
Bottom: Police intercepts speeder at point xI. Speeder moving at vS0 and police car moving at VPI
Note: Drawing not to scale.

Part a) How long does it take for the police car to intercept the speeder?

First, let’s look at the police car’s equations of motion.

xPI = x0P + V0Pt + ½at2
since the police car starts at 0 and at rest, v0P=0 then
xP = ½at2

vPI = v0P + at
vPI = at

Now for the speeder’s car’s equations of motion.

xS = x0S + V0St + ½at2
x0 = 0 and the speeder is not accelerating, a = 0, therefore
xS = v0St

vS = v0S + at
vS = V0S
vS = 120 km/hr
Convert to m/s since our acceleration is in m/s2 and it probably won’t take hours for the police car to catch up.

vS = 33.33 m/s

The two vehicles were in the same position at the very beginning of the chase at x = 0. We need to find where that happens again. This will happen when xPI = xS.

From above:
xPI = ½at2
xS = v0St

Set them equal to each other.

½at2 = v0St

This quadratic equation has two solutions. The first is at x = 0 m. Divide both sides by t to get the other.

½at = v0S

Solve for t

constant acceleration math set 1

using a = 2.5 m/s2 and v0S = 33.33 m/s

constant acceleration math set 1
t = 26.66 s

It takes 26.66 seconds for the police car to catch up and intercept the speeder.

Part b) How far did the police car travel before catching up to the speeder?
Now that we know the time, we can find the distance. From the police car’s position equation above:

xPI = ½at2
xPI = ½(2.5 m/s2)(26.66)2
xPI = 888.44 m

The police car travelled 888.44 m before it intercepted the speeder.

Part c) How fast was the police car travelling when it intercepts the speeder?
Again, using the time and the police car’s velocity equation from above:

vPI = at
vPI = (2.5 m/s2)(26.66 s)
vPI = 66.65 m/s

The police car was travelling at 66.65 m/s when it intercepted the speeder. If you convert it to km/hr, the speed of the police car is 239.94 hm/hr. Talk about speeding vehicles.

For another example of constant acceleration motion, check out Equations of Motion – Constant Acceleration Example Problem. This problem shows how to find details about the position, velocity and acceleration of a breaking vehicle.