Last updated on November 21st, 2021 at 10:07 am

In this post, we solve a few selected *Numerical problems on collisions of objects in two dimensions*. We will solve the collision problems using axial-vector components and the **law of conservation of momentum**.

Conservation of momentum in all closed systems is valid, regardless of the directions of the objects before and after they collide.

Most objects are not confined to a single line, like trains on a rail. Rather, many objects, like billiard balls or cars, can move in two dimensions. Conservation of momentum for these objects can also be calculated.

Momentum is a vector and collisions of objects in two dimensions can be represented by *axial-vector components*.

To review axial components, revisit posts on Resolving Vectors into Axial Components and Vector Addition. Here we start solving the collision numerical set.

## Numerical problems on collisions of objects in two dimensions

**Example Problem #1: A 2.0 kg ball, A, is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, B, also with a mass of 2.0 kg. After the collision, ball A moves off at 30° south of west while ball B moves off at 60° north of west. Find the velocities of both balls after the collision.**

**Solution:** Since ball B is stationary before the collision, then the total momentum before the collision is equal to the momentum of ball A.

The momentum of ball A * before the collision* is shown in

**red**below, and can be calculated to be:

p = mv = (2.00 kg)(5.00 m/s) = 10.0 kg m/s west

Since momentum is conserved in this collision, **total momenta after the collision** = total momenta before the collision = **10.0 kg m/s west.**

In other words, the* vector sum* of the momenta of balls A and B

**must be 10.0 kg m/s west.**

*after the collision*Now as after collision both A and B have their individual momentum, we can get those by resolving their vector sum value (10.0 kg m/s west), according to their directions after the collision, as directed by the problem statement (also see figure 1 above).

Momentum of A after the collision = P_{Afinal} = (10.0 kg m/s)(cos30) = (10.0 kg m/s)(0.866) = 8.66 kg m/s

And, Momentum of B after the collision =P_{Bfinal} = (10.0 kg m/s)(cos60) = (10.0 kg m/s)(0.500) = 5.00 kg m/s

To find the *final velocities* of the two balls, we *divide the momentum of each by its mass*.

Therefore, final velocity of A = V_{A} = P_{Afinal} / mass of A = (8.66/2) m/s = 4.3 m/s

and final velocity of B = V_{B} = P_{Bfinal} / mass of B = (5/2) m/s = 2.5 m/s .

**Example Problem #2: A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s. The two cars stick together after the collision. What is the speed and direction of the two cars after the collision?**

**Solution: **

Before collision Northward momentum = (1325 kg)(27.0 m/s) = 35800 kg m/s

Before collision Eastward momentum = (2165 kg)(17.0 m/s) = 36800 kg m/s

Hence, Magnitude of **Net momentum** **before collision** = R =√[(35800)^{2}+(36800)^{2}] = 51400 kg.m/s

Direction of Net momentum before collision: θ = tan^{-1} (35800/36800) = 44° north of east

As per the law of conservatin of momentum, the momentum after collision = momentum before collision.

Hence, after collision momentum of the two cars (that got sticked after collision) = 51400 kg.m/s in the direction 44° north of east Therefore, velocity of two cars in combined condition after the collision = momentum/mass = 51400/(1325+2165) m/s = 51400/3490 m/s = 14.7 m/s @ 44° N of E

**Example Problem #3:**

A 6.00 kg ball, A, moving at velocity 3.00 m/s due east collides with a 6.00 kg ball, B, at rest. After the collision, A moves off at 40.0° N of E and ball B moves off at 50.0° S of E.

a. What is the momentum of A after the collision?

b. What is the momentum of B after the collision?

c. What are the velocities of the two balls after the collision?

**Solution: **

More Explanation coming soon.

## Related study (collisions)

**Collisions – definitions, types, sample numerical**

**Numerical problems – collisions**

**Numerical problems – 2D collisions**