Question

Mass A moving at 12 m/s hits Mass B which is initially at rest. Mass A is 7 kg while Mass B is 9 kg. What will their velocities be after (a) an elastic collision and (b) perfectly inelastic collision?

Answer 1

The total momentum in any closed system will remain constant. When two or more objects collide, the collision does not change the total momentum of the two objects. Whatever momentum is lost by one object in the collision is gained by the other. The total momentum of the system is conserved.

There are three types of collisions:

1. Elastic

2. Completely Inelastic *

3. Inelastic ** 

When an object hits another object and bounce off each other, the collision would be Elastic:

$Formula: m_1\times u_1+ m_2\times u_2 = m_1\times v_1 + m_2\times v_2;$ _.

Where $u_1, u_2$ are the velocity before the collision and $v_1,v_2$ are the velocities after the collision. 

When an object hits another object and sticks together, the collision would be totally inelastic. Since the objects stick together, they have the same final velocity.

$Formula: m_1\times u_1+ m_2\times u_2 = (m_1 + m_2)\times v_F;$

Where v (u₁and u₂) is the initial velocity of the objects and v_F is the final velocity of the objects stuck together.

In this exapmle, mass of A is m₁ = 7 kg, mass of B is m₂ = 9 kg, velocity of A is u₁ = 12 $\frac{m}{s}$ and velocity of B is u₂ = 0 $\frac{m}{s}$ .

So then can compute (a) and (b) cases in the example:

(a).

In an elastic collision, both momentum and kinetic energy are conserved. Consider particles 1 and 2 with masses m₁, m₂, and velocities u₁, u₂ before collision, $v_1,v_2$  after collision. The conservation of the total momentum before and after the collision is expressed by:

$m_1\times u_1= m_1\times v_1 + m_2\times v_2;$

Likewise, the conservation of the total kinetic energy is expressed by:

$\frac{1}{2} \times m_1\times u^2_1= \frac{1}{2} \times m_1\times v^2_1 + \frac{1}{2} \times m_2\times v^2_2;$

These equations may be solved directly to find $v_1,v_2$  when  $u_1, u_2$  are known:

$v_1=\frac{m_1-m_2}{m_1+m_2}\times u_1=\frac{7-9}{7+9}\times 12=-\frac{1}{8}\times12=-\frac{3}{2};$

$v_1=\frac{2\times m_1}{m_1+m_2}\times u_1=\frac{2\times7}{7+9}\times 12=\frac{7}{8}\times12=\frac{21}{2}$ $=10.5;$

(b).

Question: What will their velocities be after an elastic collision?

Answer: Formula: $Formula: m_1\times u_1= (m_1+ m_2)\times v_F;$

should find $v_F$ .

$v_F=\frac{m_1\times u_1}{m_1\times m_2}=\frac{7 \times 12}{7 + 9 }=\frac{84}{16}=\frac{21}{4}=5.25;$