Collision

Collision is the interaction of two objects in which the kineticenergy is transferred from one object to the other, yet the total momentum is conserved. That means, the total momentum is equal before and after collision. Collision has two types - Elastic collision and Inelastic collision.

Elastic collision happens when the kinetic energy of the system after collision is conserved. That is, the total kinetic energy before the two objects collide is the same after the collision. Two objects collide elastically if they separate from each other after the collision and still maintain their shape - collision of two balls, wrestlers' body is the same after the fight, the shuttlecock flew away after it was hit by the badminton racket. Perfectly elastic collision occurs when two objects separate with different velocities after collision but the kinetic energy remains unchaged, thus, the kinetic energy is conserved.

Inelastic collision causes a decrease of kinetic energy of the system after the collision. Deformity of the objects is one way of detemining this type of collision - a wrecked car after collision, a coin get stucked on the clay and the softdrink can crashed after a big stone is thrown to it. Perfectly inelastic collision happens when the two objects got stuck with each other and move as one after the collision. Kinetic energy is not conserved in this type of collision.

The collision of two objects, A and B can be mathematically expressed as

$$m_Av_A+m_Bv_B=m_Av'_A+m_Bv'_B$$

where  $$m_A \; \text{and}\; m_B=\text{mass of object A and B, respectively}\\ v_A \; \text{and}\; v_B=\text{velocities before collision}\\ v'_A \; \text{and}\; v'_B=\text{velocities after collision}$$

In terms of the kinetic energy conserved or changed after the collision, the equation will become

$$\frac 12[m_A(v_A)^2+m_B(v_B)^2]=\frac 12[m_A(v'_A)^2+m_B(v'_B)^2]$$.

Example 1.

Consider two objects A and B collide with each other.

Before collision:

Object A with mass $$m=6.0\;kg$$ is at rest while object B with mass $$m=8.0\;kg$$ is moving with velocity $$v=9.0\;m/s$$.

After collision:

Object A is moving at $$v=8\;m/s$$ while object B is at rest.

Compute for the total kinetic energy of the system before and after collision. Determine the type of collision of the two objects.

Given:

$$m_A=6.0\;kg \quad\quad\quad\quad m_B=8.0\;kg\\ v_A=0\;m/s \quad\quad\quad\quad v_B=9.0\;m/s\\ v'_A=8\;m/s \quad\quad\quad\quad v'_B=0\;m/s$$

Solution 1.

Solve for the total kinetic energy before collision.

$$KE=\frac 12[m_A(v_A)^2+m_B(v_B)^2]\\ \quad \;\; = \frac 12[6.0\;kg(0\;m/s)^2+8.0\;kg(9.0\;m/s)^2]\\ \quad \;\; = \frac 12[0\;J+648\;J]\\ \quad\;\;=324\;J.$$

Solution 2:

The total kinetic energy after collision is:

$$KE=\frac 12[m_A(v'_A)^2+m_B(v'_B)^2]\\ \quad\;\;=\frac 12[6.0\;kg(8.0\;m/s)^2+8.0\;kg(0\;m/s)^2]\\ \quad\;\;=\frac 12[384\;J+0\;J]\\ \quad\;\;=192\;J.$$

The total kinetic energy after the collision decreased however, the two objects separate from each other after the collision, therefore we can say that the collision is partially elastic and partially inelastic. Elastic since they separate from each other and inelastic since the kinetic energy is not conserved.

Example 2.

A $$4,535-kg$$ truck moving with velocity of $$35\;m/s$$ collided into a $$970-kg$$ car at rest at the side of the road. Because of great impact, the truck and the car get stucked together and move at $$10\;m/s$$. Calculate the total kinetic energy before and after collision and tell what type of collision happened to the truck and car.

Given:

$$m_T=4,535\;kg \quad\quad\quad\quad m_c=970\;kg\\ v_T=35\;m/s \quad\quad\quad\quad\quad v_c=0\;m/s\\ v'=10\;m/s$$

Solution 1.

Before collision:

$$KE=\frac 12[m_T(v_T)^2+m_c(v_c)^2]\\ \quad \;\; = \frac 12[4,535\;kg(35\;m/s)^2+970\;kg(0\;m/s)^2]\\ \quad\;\;=2.8 \times 10^6\;J.$$

Solution 2.

After collision:

$$KE=\frac 12[m_T(v')^2+m_c(v')^2]\\ \quad\;\;=\frac 12[4,535\;kg(10.0\;m/s)^2+970\;kg(10\;m/s)^2]\\ \quad\;\;=2.8\times 10^5\;J.$$

The total kinetic energy is decreased and the two objects stick together and move as one after the collision, thus the situation is an example of perfectly inelastic collision.

Example 3.

A table tennis player is about to serve the $$2.7-g$$ ball which is at rest. If his paddle with a mass of $$80-g$$ makes the ball moving at $$30\;m/s$$ after it hit the ball, what is the initial velocity of the paddle?Assume the paddle starts from rest and the collision is perfectly elastic.

Given:

$$m_p=80\;g=0.08\;kg \quad\quad\quad\quad\quad m_b=2.7\;g=0.0027\;kg\\ v'_p=0\;m/s \quad\quad\quad\quad\quad\quad\quad\quad\quad\;\; v_b=0\;m/s\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;\; v'_b=30\;m/s$$

Solution 1:

After collision:

$$KE=\frac 12[m_p(v'_p)^2+m_b(v'_b)^2]\\ \quad\;\;=\frac 12[0.08\;kg(0\;m/s)^2+0.0027\;kg(30\;m/s)^2]\\ \quad\;\;=1.2\;J.$$

Since the collision is perfectly elastic, kinetic energy is conserved. Thus, the kinetic energy before and after collision is equal.

$$KE=\frac 12[m_p(v_p)^2+m_b(v_b)^2]\\ 1.2\;J= \frac 12[0.08\;kg(v_p)^2+0.0027\;kg(0\;m/s)^2]\\ 1.2\;J= \frac 12[0.08\;kg(v_p)^2]\\ 1.2\;J= 0.04\;kg(v_p)^2$$

$$(v_p)^2=\frac{1.2\;J}{0.04\;kg}=30\;m^2/s^2$$

Therefore, the velocity of the paddle before collision is

$$v_p=\sqrt{30\;m^2/s^2}=5.48\;m/s.$$