Inelastic Collision Questions

1.0Inelastic Collision Definition

• An inelastic collision is a type of collision in which the total kinetic energy of the system is not conserved, even though momentum is conserved. 
• In an inelastic collision, a portion of the initial kinetic energy is converted into heat, sound, or the deformation of the colliding objects.  
• The extent of energy loss varies, with perfectly inelastic collisions being a specific case in which the colliding objects stick together and move as one combined mass after the collision.
• Coefficient of Restitution for inelastic collisions is 0<e<1
• Coefficient of Restitution for perfectly inelastic collisions e=0

2.0Formula used For Elastic and Inelastic Collision

1. Elastic Collision

• $v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+(\frac{2m_2}{m_1+m_2})u_2$
• $v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+(\frac{2m_1}{m_1+m_2})u_1$
• $e=\frac{v_2-v_1}{u_1-u_2}=1$

2. Inelastic Collision

• $v=\frac{m_1}{m_1+m_2}{u}_1$
• $\Delta K=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{u}_1^2$

3.0Solved Examples

Q-1. A body of 2 Kg mass having velocity 3 m/s collides with a body of 1 kg mass moving with a velocity of 4 m/s in the opposite direction. After collision both bodies stick together and move with a common velocity. Find the velocity?

Solution:

$v_{\text{system}}=\frac{m_1{v}_1-m_2{v}_2}{m_1+m_2}=\frac{2\times 3-1\times 4}{3}=\frac{2}{3}\text{m/s}$

Q-2.A particle with mass  m and velocity v collides with a stationary particle of mass 2m, and they stick together upon impact. What is the resulting speed of the combined mass after the collision?

Solution:

$$\begin{gathered} \text{On applying conservation of momentum} \\ {P}_i=P_f \\ m\times v-2m\times 0=(m+2m)V_{\text{System}} \\ mv=3m{V}_{\text{System}} \\ {V}_{\text{System}}=\frac{v}{3}\text{m/s} \end{gathered}$$

Q-3.A bullet of mass m travelling with a velocity u strikes a stationary block of mass m on a smooth surface. The collision is completely inelastic. We need to determine the common velocity after the collision and calculate the fractional loss in kinetic energy.

Solution:

$$\begin{gathered} \text{On applying momentum conservation} \\ mu+M\times 0=(M+m)v \\ \Rightarrow v=\frac{mu}{m+M} \\ \text{Loss in K.E, }\Delta K=\frac{1}{2}\frac{Mm}{m+M}(u-0{)}^2 \\ \text{Incident K.E, }\Delta K_i=\frac{1}{2}m{u}^2 \\ \text{Fractional Loss in K.E, } \\ \frac{\Delta K}{K_I}=\frac{M}{M+m} \end{gathered}$$

Q-4.A cart A, with a mass of 50 kg, is traveling at a speed of 20 km/h when it collides with a lighter cart B, which has a mass of 20 kg and is moving towards cart A at a speed of 10 km/h. After the collision, the two carts stick together. We need to find the speed of the combined mass after the collision.

Solution:

This is an example of an inelastic collision since the carts are moving toward each other, meaning their momenta have opposite signs. Let the common speed after the collision be V. By using principle of conservation of momentum, we can express this relationship as follows,

50 ✕ 20-20 ✕ 10=70 ✕ V

$V=\frac{80}{7}\text{km/h}$

Q-5.After a perfectly inelastic collision between two similar balls moving at the same speed but in different directions, the speed of the total mass is half of the initial speed. We need to determine the angle between the two balls before the collision.

Solution:

$$\begin{gathered} \text{Conservation of Momentum is used and }\theta \text{is the required angle} \\ {p}^2=p_1^2+p_2^2+2p_1{p}_2\text{Cos}\theta \\ {\left\{2m\left(\frac{v}{2}\right)\right\}}^2=(mv{)}^2+(mv{)}^2+2(mv)(mv)\text{Cos}\theta \\ 1=1+1+2\text{Cos}\theta \\ \text{Cos}\theta =-\frac{1}{2} \\ \theta =120^{\circ } \end{gathered}$$