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In this Physics tutorial, you will learn:

- How to calculate momentum in two dimensions
- The same for impulse in two dimensions
- The meaning of explosions as the inverse process of inelastic collisions
- How to deal with explosions in two dimensions

We have discussed so far about head-to-head collisions, in which objects move either in the same or in opposite direction. Now we will discuss about situations involving collisions that are slightly diverted from head-to-head direction, after which objects have a lateral component of motion besides the normal one. Also, we will explain what happens during collisions and what are the common things and differences between collisions and explosions.

A head-to-head collision (as discussed in the previous two articles), occurs only when the motion of objects occurs according the line that connects their respective centres of mass, as shown in the figure below.

The figure shows an upper view of two balls moving towards each other. Even after the collision, they will continue moving according the line that connects their centres of mass, either in the same or in opposite direction they had prior to collision, depending on their respective masses and initial velocities.

But when objects are not moving according the line that connects their centres of mass but slightly diverted from it, as shown in the figure,

the collisions will not be head-to-head anymore. As a result, objects will move on other directions as shown below.

As a result, the study of momentum and impulse for this kind of collision involves two directions: x- and y-direction as the movement of the two objects after the collision is a combination of these two basic directions.

The approach is the same as for momentum and impulse in one direction discussed earlier. The only difference is that we must write the known equations for each direction separately.

Thus, for elastic collisions we have:

and for inelastic collisions, we have

Normally we don't have to use all the four above equations in the same exercise. Even if the initial motion is a combination of the two basic directions, we can rotate the axes to make one of them fit to the direction of motion, just as we did when studying the forces acting at an object on an inclined plane (tutorial "Newton's Second Law of Motion"). Below, an illustration showing two objects moving towards each other but not in the head-to-head direction is shown.

The angles formed by object's trajectory to the horizontal direction (if not given), are calculated by using trigonometric functions, as usual.

A 200 g ball moving at 4 m/s collides with a 400 g ball initially at rest. After the collision, the first ball moves at 3 m/s and at 37° to the original direction as shown in the figure.

**Calculate:**

- The components of velocity of the 400 g ball after the collision.
- The magnitude of velocity of the 400 g ball after the collision.
- The angle formed by the 400 g object to the horizontal direction after the collision.

Take cos 37° = 0.8 and sin 37° = 0.6.

First, let's write the clues. Since the first ball initially moves in the x-direction only and the second object is at rest, we have:

m_{1} = 200 g = 0.2 kg

m_{2} = 400 g = 0.4 kg

v_{01(x)} = 4 m/s

v_{01(y)} = 0

v_{02(x)} = 0

v_{02(y)} = 0

θ_{1} = 37°

v_{1} = 3 m/s

v_{2(x)} = ?

v_{2(y)} = ?

v_{2} = ?

θ_{2} = ?

m

v

v

v

v

θ

v

v

v

v

θ

Also, we have:

v_{1(x)} = v_{1} × cos θ_{1} = 3 × 0.8 = 2.4 m/s

v_{1(y)} = v_{1} × sin θ_{1} = 3 × 0.6 = 1.8 m/s

v

**a.** Let's calculate the components of the 400 g object's velocity after the collision.

m_{1} × v*⃗*_{01(x)} + m_{2} × v*⃗*_{02(x)} = m_{1} × v*⃗*_{1(x)} + m_{2} × v*⃗*_{2(x)}

0.2 × 4 + 0.4 × 0 = 0.2 × 2.4 + 0.4 × v*⃗*_{2(x)}

0.8 = 0.48 + 0.4 × v*⃗*_{2(x)}

v*⃗*_{2(x)} = *0.8 - 0.48**/**0.4 * = 0.8 m/s

m_{1} × v*⃗*_{01(y)} + m_{2} × v*⃗*_{02(y)} = m_{1} × v*⃗*_{1(y)} + m_{2} × v*⃗*_{2(y)}

0.2 × 0 + 0.4 × 0 = 0.2 × 1.8 + 0.4 × v*⃗*_{2(y)}

0 = 0.36 + 0.4 × v*⃗*_{2(y)}

0.4 × v*⃗*_{2(y)} = -0.36

v*⃗*_{2(y)} = *-0.36**/**0.4* = -0.9 m/s

0.2 × 4 + 0.4 × 0 = 0.2 × 2.4 + 0.4 × v

0.8 = 0.48 + 0.4 × v

v

m

0.2 × 0 + 0.4 × 0 = 0.2 × 1.8 + 0.4 × v

0 = 0.36 + 0.4 × v

0.4 × v

v

These results mean the second ball will move at 0.8 m/s right and 0.9 m/s down.

**b.**The magnitude of the second object's velocity after the collision is

|v*⃗*_{2}| = √**v**^{2}_{2(x)} + v^{2}_{2(y)}

= √**0.8**^{2} + (-0.9)^{2}

= √**0.64 + 0.81**

= √**1.45**

= 1.20 m/s

= √

= √

= √

= 1.20 m/s

**c.**We can use any of trigonometric equations to find the angle θ_{2} formed by the second ball to the horizontal direction. We have:

tan θ_{2} = *sin θ*_{2}*/**cos θ*_{2}

=*v**⃗*_{2(y)}*/**v**⃗*_{2(x)}

=*-0.9**/**0.8*

= -1.125

=

=

= -1.125

This tangent value corresponds to two angles: -48.370 and 131.630. We will take the first angle as it fits the description (the velocity lies right-down).

The final figure that includes everything we found so far, is shown below.

We can use the same approach as above for the impulse in two dimensions. Thus, since there is a single collision, there is also a single force exerted by an object to the other (and a single reaction force as well). However, this force extends in two directions and therefore, we must use two equations for the impulse - one for each direction.

We have:

J*⃗*_{1 - 2} = F*⃗*_{1 - 2} × ∆t = m_{1} × v*⃗*_{01}-m_{1} × v*⃗*_{1}

and

J*⃗*_{2 - 1} = F*⃗*_{2 - 1} × ∆t = m_{2} × v*⃗*_{02}-m_{2} × v*⃗*_{2}

Obviously, we need to calculate only one of the above impulses as J*⃗*_{1 - 2} and J*⃗*_{2 - 1} are equal and opposite. However, we must write it in components. For example, if we chose to work out the first impulse (J*⃗*_{1 - 2}), we must write

J*⃗*_{1 - 2(x)} = F*⃗*_{1 - 2(x)} × ∆t = m_{1} × v*⃗*_{01(x)}-m_{1} × v*⃗*_{1(x)}

and

J*⃗*_{1 - 2(y)} = F*⃗*_{1 - 2(y)} × ∆t = m_{1} × v*⃗*_{01(y)}-m_{1} × v*⃗*_{1(y)}

For illustration, let's calculate the force of collision for the situation described in the example of the previous paragraph. Let's suppose the collision lasts for 0.05 s and we are required to find the collision force.

We have:

F*⃗*_{1 - 2(x)} × ∆t = m_{1} × v*⃗*_{01(x)} - m_{1} × v*⃗*_{1(x)}

F*⃗*_{1 - 2(x)} × 0.05 = 0.2 × 4 - 0.2 × 2.4

F*⃗*_{1 - 2(x)} × 0.05 = 0.8 - 0.48

F*⃗*_{1 - 2(x)} × 0.05 = 0.32

F*⃗*_{1 - 2(x)} = *0.32**/**0.05* = 6.4 N

F*⃗*_{1 - 2(y)} × ∆t = m_{1} × v*⃗*_{01(y)} - m_{1} × v*⃗*_{1(y)}

F*⃗*_{1 - 2(y)} × 0.05 = 0.2 × 0 - 0.2 × (-0.9)

F*⃗*_{1 - 2(y)} × 0.05 = 0.18

F*⃗*_{1 - 2(y)} = *0.18**/**0.05* = 3.6 N

F

F

F

F

F

F

F

F

Hence, the force of collision is

|F*⃗*_{1 - 2} | = √**F***⃗*_{1 - 2(x)}^{2} + F*⃗*_{1 - 2(y)}^{2}

= √**6.4**^{2} + 3.6^{2}

= √**41 + 13**

= √**54**

= 7.3 N

= √

= √

= √

= 7.3 N

The same result is obtained by considering the values of the second ball as well. (Try it!).

Explosions are the reverse phenomenon of inelastic collisions. Thus, in inelastic collisions the kinetic energy converts into other form after the impact, while in explosions, the energy stored in the object converts into kinetic energy, making it move faster than before. Therefore, the equation of explosions is the inverse of that used in inelastic collisions.

M × v*⃗*_{0} = m_{1} × v*⃗*_{1} + m_{2} × v*⃗*_{2}

where M is the mass of the original object, m_{1} and m_{2} are the masses of the respective pieces after the explosion, v_{0} is the initial velocity of the original object before the explosion, v_{1} and v_{2} are the velocities of the pieces after the explosion.

The simplest case of explosions is when the original object explodes and splits in two pieces which move in opposite directions as shown in the figure.

In this case, we have only one dimension involved and therefore, a single equation to write (the equation of explosions written above). Let's consider an example to explain this point.

A 5 kg ball is moving initially due right at 1 m/s. The ball explodes due to an internal process and as a result, it divides in two pieces at ratio 3:2 between them. After explosion, the smallest piece contine moving in the original direction at 15 m/s. What is the velocity (and the direction) of the largest piece after the explosion?

First let's work out the masses of the two pieces after the explosion. We have the ratio

m_{1} ∶ m_{2} = 3 ∶ 2

This means m_{1} = 3x and m_{2} = 2x (where x is a constant).

Also, we have m_{1} + m_{2} = 5 kg.

Therfore, we can write

3x + 2x = 5 kg

5x = 5 kg

x = 1 kg

5x = 5 kg

x = 1 kg

Thus,

m_{1} = 3 × 1 kg = 3 kg

and

m_{2} = 2 × 1 kg = 2 kg

Now, applying the equation

M × v*⃗*_{0} = m_{1} × v*⃗*_{1} + m_{2} × v*⃗*_{2}

where M = 5 kg, v*⃗*_{0} = 1 m/s and v*⃗*_{1} = 15 m/s, we find for the velocity v*⃗*_{2} of the second piece after the explosion,

5 × 1 = 3 × v*⃗*_{1} + 2 × 15

5 = 3 × v*⃗*_{1} + 30

- 25 = 3 × v*⃗*_{1}

v*⃗*_{2} = - *25**/**3* *m**/**s* ≈ - 8.33 m/s

5 = 3 × v

- 25 = 3 × v

v

This result means the larger piece will move at 8.33 m/s in the opposite direction as it was before the explosion, i.e. due left.

In explosions in two dimensions, we use the same approach as for momentum and impulse in two dimensions, i.e. we write the equation of momentum conservation for each direction separately. We have

M × v*⃗*_{0x} = m_{1} × v*⃗*_{1}x + m_{2} × v*⃗*_{2}x + m_{3} × v*⃗*_{3x} + …

M × v*⃗*_{0}y = m_{1} × v*⃗*_{1}y + m_{2} × v*⃗*_{2}y + m_{3} × v*⃗*_{3y} + …

For illustration, let's consider an example where an object explodes and divides in three pieces.

A 12 kg object moving at 3 m/s due right - as shown in the figure below - explodes in three pieces, which have a ratio of 2:1:3 between them.

After the explosion, the first piece moves at 8 m/s up, the second piece (the smallest) moves at 40 m/s due right (in the same direction it was moving before the explosion).

**Calculate:**

- The velocity components of the largest piece after the explosion
- Magnitude of velocity of the largest piece after the explosion
- The angle formed by the largest piece to the left-right direction after the explosion
- What happens with the kinetic energy of the object after the explosion? Explain.

First, let's calculate the masses of the three pieces after the explosion. Giving that the total mass of the original object was M = 12 kg, we have

m_{1} ∶ m_{2} ∶ m_{3} = 2 ∶ 1 ∶ 3

and

m_{1} + m_{2} + m_{3} = 12 kg

From the first equation, we obtain

m_{1} = 2x

m_{2} = 1x

m_{3} = 3x

m

m

where x is a constant. Substituting them in the second equation, we obtain

2x + 1x + 3x = 12 kg

6x = 12 kg

x = 2 kg

6x = 12 kg

x = 2 kg

Therefore,

m_{1} = 2x = 2 × 2 kg = 4 kg

m_{2} = 1x = 1 × 2 kg = 2 kg

m_{3} = 3x = 3 × 2 kg = 6 kg

m

m

We can take the let-to-right as positive of x-direction and down-to-up as positive of y-direction (as usual). Therefore, the other clues of this problem are:

v_{0x} = 3 m/s

v_{0y} = 0

v_{1x} = 0

v_{1y} = 8 m/s

v_{2x} = 40 m/s

v_{2y} = 0

v_{3x} = ?

v_{3y} = ?

v

v

v

v

v

v

v

Applying the law of conservation of momentum for two dimensions,

M × v*⃗*_{0x} = m_{1} × v*⃗*_{1x} + m_{2} × v*⃗*_{2x} + m_{3} × v*⃗*_{3x}

M × v*⃗*_{0y} = m_{1} × v*⃗*_{1y} + m_{2} × v*⃗*_{2y} + m_{3} × v*⃗*_{3y}

we obtain after the substitutions

12 × 3 = 4 × 0 + 2 × 40 + 6 × v*⃗*_{3x}

-44 = 6 × v*⃗*_{3x}

v*⃗*_{3x} = -7.3 m/s

12 × 0 = 4 × 8 + 2 × 0 + 6 × v*⃗*_{3y}

-32 = 6 × v*⃗*_{3x}

v*⃗*_{3y} = -5.3 m/s

-44 = 6 × v

v

12 × 0 = 4 × 8 + 2 × 0 + 6 × v

-32 = 6 × v

v

This means the third and largest piece will move at 7.3 m/s left and 5.3 m/s down after the explosion as shown in the figure below.

**b.** |v*⃗*_{3} | = ?

Magnitude of the velocity of the largest piece after explosion is

|v*⃗*_{3} | = √**v**^{2}_{3x} + v^{2}_{3y}

= √**(-7.3)**^{2} + (-5.3)^{2}

= √**53.3 + 28.1**

= √**82**

= 9.1 m/s

= √

= √

= √

= 9.1 m/s

**c.** We have to calculate the size of the wide angle θ formed by the third piece velocity vector and the left-to-right direction. We have:

θ = tan^{-1} *v*_{3y}*/**v*_{3x}

= tan^{-1} *-5.3 m/s**/**-7.3 m/s*

= 36° or 180° + 36°

= 216°

= tan

= 36° or 180° + 36°

= 216°

We consider only the second value as the angle θ is wide as shown in the figure.

**d.** The initial velocity of the object before explosion was directed only according the x-direction. Therefore, v_{0} = v_{0x} = 3 m/s.

Kinetic energy of the object before the explosion therefore was

KE_{0} = *M × v*^{2}_{0}*/**2* = *12 × 3*^{2}*/**2* = 54 J

On the other hand, giving that v_{1} = v_{1y} = 8 m/s, v_{2} = v_{2x} = 40 m/s and v_{3} = 13.6 m/s (as foud earlier), the kinetic energy of each piece after the explosion is

KE_{1} = *m*_{1} × v^{2}_{1}*/**2* = *4 × 8*^{2}*/**2* = 128 J

KE_{2} = *m*_{1} × v^{2}_{2}*/**2* = *2 × 40*^{2}*/**2* = 1600 J

KE_{3} = *m*_{1} × v^{2}_{3}*/**2* = *6 × 82**/**2* = 246 J

KE

KE

Therefore, the total kinetic energy KE of the system after the explosion is

KE = KE_{1} + KE_{2} + KE_{3}

= 128 J + 1600 J + 246 J

= 1974 J

= 128 J + 1600 J + 246 J

= 1974 J

Obviously, this value is much gretaer than the initial kinetic energy of the object before expoding. This means some of the stored (idle) energy of the object has been activated and it was converted into kinetic energy during the explosion. Therefore, there is no break in the energy conversion law discusses in the previous articles.

When objects are not moving according the line that connects their centres of mass but slightly diverted from it, the collision is not head-to-head. As a result, objects will move in more than one direction and the study of momentum and impulse for this kind of collision involves two directions: x- and y-direction as the movement of the two objects after the collision is a combination of these two basic directions.

The approach is the same as for momentum and impulse in one direction discussed earlier. The only difference is that we must write the known equations for each direction separately.

Thus, for elastic collisions we have:

and for inelastic collisions, we have

We can use the same approach as above for the impulse in two dimensions. Thus, since there is a single collision, there is also a single force exerted by an object to the other (and a single reaction force as well). However, this force extends in two directions and therefore, we must use two equations for the impulse - one for each direction.

We have:

J*⃗*_{1 - 2} = F*⃗*_{1 - 2} × ∆t = m_{1} × v*⃗*_{01} - m_{1} × v*⃗*_{1}

and

J*⃗*_{2-1} = F*⃗*_{2-1} × ∆t = m_{2} × v*⃗*_{02} - m_{2} × v*⃗*_{2}

Obviously, we need to calculate only one of the above impulses as J*⃗*_{1 - 2} and J*⃗*_{2-1} are equal and opposite. However, we must write it in components. For example, if we chose to work out the first impulse (J*⃗*_{1 - 2}), we must write

J*⃗*_{1 - 2(x)} = F*⃗*_{1 - 2(x)} × ∆t = m_{1} × v*⃗*_{01(x)} - m_{1} × v*⃗*_{1(x)}

and

J*⃗*_{1 - 2(y)} = F*⃗*_{1 - 2(y)} × ∆t = m_{1} × v*⃗*_{01(y)} - m_{1} × v*⃗*_{1(y)}

Explosions are the reverse phenomenon of inelastic collisions. Thus, in inelastic collisions the kinetic energy converts into other form after the impact, while in explosions, the energy stored in the object converts into kinetic energy, making it move faster than before. Therefore, the equation of explosions is the inverse of that used in inelastic collisions.

M × v*⃗*_{0} = m_{1} × v*⃗*_{1} + m_{2} × v*⃗*_{2}

where M is the mass of the original object, m_{1} and m_{2} are the masses of the respective pieces after the explosion, v0 is the initial velocity of the original object before the explosion, v_{1} and v_{2} are the velocities of the pieces after the explosion.

The simplest case of explosions is when the original object explodes and splits in two pieces, which move in opposite directions. In this case, we have only one dimension involved and therefore, a single equation to write (the equation of explosions written above).

In explosions in two dimensions, we use the same approach as for momentum and impulse in two dimensions, i.e. we write the equation of momentum conservation for each direction separately. We have

M × v*⃗*_{0x} = m_{1} × v*⃗*_{1x} + m_{2} × v*⃗*_{2x} + m_{3} × v*⃗*_{3x} + …

M × v*⃗*_{0y} = m_{1} × v*⃗*_{1y} + m_{2} × v*⃗*_{2y} + m_{3} × v*⃗*_{3y} + …

The total kinetic energy of a system after explosion is much gretaer than its initial kinetic energy when the object had not exploded yet. This means some of the stored (idle) energy of the object has been activated and it was converted into kinetic energy during the explosion. No break in the rules of energy conversion law occurs here.

**1)** A ball moving at 6 m/s collides with an identical ball at rest. As a result, the first ball moves at 4 m/s at 530 to the original direction as shown in the figure.

What is the magnitude of velocity v of the second object after the collision?

- 3.6 m/s
- 3.2 m/s
- 4.8 m/s
- 6.0 m/s

**Correct Answer: C**

**2)** A 50 g bullet hits a 20 kg metal plate at 300 to its surface as shown in the figure. The initial velocity of the bullet is 400 m/s. As a result, the metal plate moves back at 20 cm/s. If the collision lasts for 0.01 s, what is the magnitude of force exerted by thje bullet on the metal plate? Take cos 300 = 0.86, sin 300 = 0.5.

- 800 N
- 344 N
- 464 N
- 400 N

**Correct Answer: A**

**3)** A 10 kg moving at 2 m/s explodes in two pieces: 8 kg and 2 kg. The 8 kg piece turns back at 3 m/s. What is the velocity of the 2 kg piece after explosion?

- 44 m/s in the original direction
- 22 m/s in the original direction
- 22 m/s in the opposite direction
- 2 m/s in the original direction

**Correct Answer: B**

We hope you found this Physics tutorial "Momentum and Impulse in Two Dimensions. Explosions." useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of "Momentum Impulse Explosions" with our Physics tutorial on Torque.

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