Momentum is another measurement of the vector. Momentum and the velocity both are in the same direction. Scientists do the calculation of the momentum by doing the multiplication of the mass of the object and the velocity of the object. It indicates how hard it would be for stopping the object. Learn the momentum formula here.

**Definition**

If we are standing at the bottom of the mountain and we are faced with the option of stopping a runaway semi-truck or stopping a runaway bicycle. Here we have probably chosen to stop the bicycle. This is because the semi-truck has more momentum whereas the bike is having less. The simple meaning of the momentum is mass in motion.

The semi-truck has more momentum than the bike because it has a lot of mass, but its speed is also more. Notably, the speed also influences the momentum. The bike has the momentum as well because it has a large speed, but because of its less mass than the truck, its momentum is less too.

**Derivation of Conservation of Momentum**

Let’s say there is a situation wherein: a truck of mass \(m_{1}\), velocity \(u_{1}\) and its momentum is \(m_{1}u_{1}\) and a car of mass \(m_{2}\), velocity \(u_{2}\) and its momentum \(m_{2}u_{2}\); are moving in the similar direction but their speeds are different. Therefore, the total momentum = \(m_{1}u_{1}\) + \(m_{2}u_{2}\).

Now let’s assume that the car and the truck collide for a short time t, their velocity changes. So now the velocity of the truck and the car is \(v_{1}\), and \(v_{2}\), respectively. However, their mass stays the same.

Hence, now the total momentum = \(m_{1}v_{1} + m_{2}v_{2}\)

Acceleration of car (a) = \((v_{2} – u_{2})\)/t

Also, f = ma

\(F_{1}\) = force exerted on the car by the truck.

\(F_{1}\) = \(m_{2}(v_{2}-u_{2})\)/t

Acceleration of the truck = \((v_{1}-u_{1})\)/t

\(F_{2}\) = \(m_{1}\)(\(v_{1}-u_{1}\))/t and \(F_{1}\) = \(-F_{2}\)

\(M_{2}\)(\(v_{2}-u_{2}\))/t = \(-m_{1}\)(\(v_{1}-u_{1}\))/t

\(m_{2}v_{2}-m_{2}u_{2}\) = \(-m_{1}v_{1}+m_{1}u_{1}\)

Or \(m_{1}u_{1}+m_{2}u_{2}\) = \(m_{2}v_{2}+m_{1}v_{1}\)

**Solved Example on Momentum Formula**

**Problem**

A 4.88-kg object that has a speed of 3.14 m/s strikes a plate of steel at an angle of 42.0º and rebounds at the similar speed and angle as shown in the figure below. What is the change (magnitude and direction) in the linear momentum of the object?

**Solution:-**

The below figure shows a 4.88-kg object with a speed of 31.4 m/s strikes a plate of steel at an angle of 42.0° and rebounds at the similar speed with the same angle which is visible geometrically.

A 4.88-kg Object with a Speed of 31.4 m/s strikes a plate of steel at an Angle of 42.0° and Rebounds at the Same Speed with the Same Angle.

The \(p_{i}\) is the initial momentum of the object when it strikes the steel plate.

To find initial momentum \(p_{i}\), substitute \(p_{i}\) for p, 4.88-kg for mass m of the object and 31.4 m/s for the velocity of the object present in the equation p = mv,

\(p_{i}\) = mv

= (4.88 kg) (31.4 m/s)

= 153 kg. m/s …… (2)

The final momentum \(p_{f}\) of the object is equal to the initial momentum \(p_{i}\) of the object, therefore,

\(p_{f}\) = 153 kg. m/s …… (3)

In the figure shown above the angle θ is:

θ = 42° + 42°

= 84°

With the use of the cosine law, the magnitude of change of the linear momentum of the object Δp will be,

Δp = √( \(p_{i}\))2 + (\(p_{f}\))2 + 2 (-\(p_{i}\))(\(p_{f}\))cos θ …… (4)

For getting the magnitude of linear momentum Δp, substitute 153 kg. m/s for \(p_{i}\), 153 kg. m/s for the \(p_{f}\) and 84° for the angle θ in the equation Δp = √( \(p_{i}\))2 + (\(p_{f}\))2 + 2 (-\(p_{i}\))(\(p_{f}\)) cos θ,

Δp = √( \(p_{i}\))2 + (pf)2 + 2 (-\(p_{i}\))(\(p_{f}\)) cos θ

= √(153 kg. m/s)2 + (153 kg. m/s)2 – 2 (153 kg. m/s)( 153 kg. m/s) cos 84°

= 205 kg. m/s

From the observation given above, we conclude that the change in the linear momentum of the object will be 205 kg. m/s and also from the figure the direction of change of the linear momentum of the object is perpendicular to the plate of steel.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…