Practice elastic collision in two dimensions and equation of inelastic collision. Learn elastic and inelastic collision formula and definition by practice problems.
Momentum Before Collision = Momentum During Collision = Momentum After Collision
\(\Rightarrow KE_{BC} = \dfrac{1}{2} m_1\,u_1^2 + \dfrac{1}{2} m_2\,u_2^2\)
\(\Rightarrow KE_{AC} = \dfrac{1}{2} m_1\,v_1^2 + \dfrac{1}{2} m_2\,v_2^2\)
\(\Rightarrow K.E_{BC} = K.E_{AC}\neq K.E_{DC}\)
\(\Rightarrow \dfrac{1}{2}m_1\,u^2_1+\dfrac{1}{2}m_2\,u^2_2=\dfrac{1}{2}m_1\,v^2_1+\dfrac{1}{2}m_2\,v^2_2\) ..........(1)
Momentum Before Collision = Momentum During Collision = Momentum After Collision
\(\Rightarrow m_1\,u_1+m_2\,u_2=m_1\,v+m_2\,v=m_1\,v_1+m_2\,v_2\) ........(2)
A \(m_1\,u_1 + m_2\,u_2 = m_1\,v+m_2\,v\)
B \(m_1\,v_1 + m_2\,v_2 = m_1\,v+m_2\,v\)
C \(\dfrac{1}{2}m_1\,u_1^2 + m_2\,u_2^2 = \dfrac{1}{2} m_1\,v^2+m_2\,v^2\)
D \(\dfrac{1}{2}m_1\,u_1^2 + \dfrac{1}{2}m_2\,u_2^2 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2}m_2\,v_2^2\)
\(K.E_{\text {After collision}} <K.E_ {\text {Before collision}} \)
and \(p_{\text {Before collision}} =p_{\text {During collision}} =p_{\text {After collision}}\)
Where \(K.E={\text {Kinetic energy}}\)
\(\Rightarrow p={\text {momentum}}\)
A \(m_1\,u_1 - m_2\,v_2 = m_1\,v +m_2\,v\)
B \(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 = \dfrac{1}{2}m_1\,v^2 +\dfrac{1}{2} m_2\,v^2\)
C \(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 < \dfrac{1}{2}m_1\,v^2_1 +\dfrac{1}{2} m_2\,v^2_2\)
D \(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 > \dfrac{1}{2}m_1\,v^2_1 +\dfrac{1}{2} m_2\,v^2_2\)
Momentum Before Collision = Momentum During Collision = Momentum After Collision
?\(\Rightarrow KE_{BC} = \dfrac{1}{2} m_1\,u_1^2 + \dfrac{1}{2} m_2\,u_2^2\)
\(\Rightarrow KE_{AC} = \dfrac{1}{2} m_1\,v_1^2 + \dfrac{1}{2} m_2\,v_2^2\)
\(\Rightarrow K.E_{BC} = K.E_{AC}\neq K.E_{DC}\)
\(\Rightarrow \dfrac{1}{2}m_1\,u^2_1+\dfrac{1}{2}m_2\,u^2_2=\dfrac{1}{2}m_1\,v^2_1+\dfrac{1}{2}m_2\,v^2_2\) ..........(1)
Momentum Before Collision = Momentum During Collision = Momentum After Collision
\(\Rightarrow m_1\,u_1+m_2\,u_2=m_1\,v+m_2\,v=m_1\,v_1+m_2\,v_2\) ........(2)
\(\Rightarrow m_1\,u_1 +m_2\,u_2 = m_1\,v_1 +m_2\,v_2 \) .......(3)
and \(\dfrac{1}{2}m_1\,u_1^2 +m_2\,u_2^2 =\dfrac{1}{2} m_1\,v_1^2 +\dfrac{1}{2}m_2\,v_2^2 \) ......(4)
Calculation of v_{1 }and v_{2}
\(\Rightarrow m_1(u_1-v_1) = m_2(u_2-v_2) \) .....(5)
and \(m_1(u_1^2-v_1^2) = m_2(u_2^2-v_2^2) \)
\(\Rightarrow m_1(u_1-v_1)(u_1 +v_1) = m_2(v_2-u_2)(v_2 +u_2)\) .........(6)
\(\Rightarrow u_1 +v_1 = v_2+u_2\)
or, \( u_1 -u_2 = v_2-v_1\)
or, \(( u_1 -u_2) = -(v_1-v_2)\)
Conclusion
Thus, in elastic collision the relative velocity of particles before collision is the negative of the relations velocity after collision.
\(\Rightarrow v_1 = \left(\dfrac{m_1-m_2}{m_1+m_2}\right)\,u_1+ \left(\dfrac{2\,m_2}{m_1+m_2}\right)\,u_2\) .......(7)
and \( v_2 = \left(\dfrac{2\,m_1}{m_1+m_2}\right)\,u_1+ \left(\dfrac{\,m_2-m_1}{m_1+m_2}\right)\,u_2\) ........(8)
then, from equation (7)
\(v_1 = u_1\)
and from equation (8),
\(\Rightarrow v_2 = 2\,u_1 - u_2\)
Conclusion
The velocity of a heavier body remains unchanged during collision.
A \(5 \,m/sec\)
B \(6\,m/sec\)
C \(11 \,m/sec\)
D \(2 \,m/sec\)
Momentum Before Collision = Momentum During Collision = Momentum After Collision
\(\Rightarrow KE_{BC} = \dfrac{1}{2} m_1\,u_1^2 + \dfrac{1}{2} m_2\,u_2^2\)
\(\Rightarrow KE_{AC} = \dfrac{1}{2} m_1\,v_1^2 + \dfrac{1}{2} m_2\,v_2^2\)
So, kinetic energy before collision is same as kinetic energy after collision.
\(\Rightarrow K.E_{BC} = K.E_{AC}\neq K.E_{DC}\)
\(\Rightarrow \dfrac{1}{2}m_1\,u^2_1+\dfrac{1}{2}m_2\,u^2_2=\dfrac{1}{2}m_1\,v^2_1+\dfrac{1}{2}m_2\,v^2_2\) ..........(1)
Since, no external force is applied during collision so, momentum of the system remains conserved.
Momentum Before Collision = Momentum During Collision = Momentum After Collision
\(\Rightarrow m_1\,u_1+m_2\,u_2=m_1\,v+m_2\,v=m_1\,v_1+m_2\,v_2\) ........(2)
\(\Rightarrow m_1\,u_1 +m_2\,u_2 = m_1\,v_1 +m_2\,v_2 \) .......(3)
and \(\dfrac{1}{2}m_1\,u_1^2 +m_2\,u_2^2 =\dfrac{1}{2} m_1\,v_1^2 +\dfrac{1}{2}m_2\,v_2^2 \) ......(4)
Calculation of \(v_1 \,\,\text{and}\,\,v_2\)
\(\Rightarrow m_1(u_1-v_1) = m_2(u_2-v_2) \) .....(5)
and \(m_1(u_1^2-v_1^2) = m_2(u_2^2-v_2^2) \)
\(\Rightarrow m_1(u_1-v_1)(u_1 +v_1) = m_2(v_2-u_2)(v_2 +u_2)\) ........(6)
\(\Rightarrow u_1 +v_1 = v_2+u_2\)
or, \( u_1 -u_2 = v_1-v_2\)
or, \(( u_1 -u_2) = -(v_1-v_2)\)
Conclusion
Thus, in elastic collision the relative velocity of particles before collision is the negative of the relations velocity after collision.
\(\Rightarrow v_1 = \left(\dfrac{m_1-m_2}{m_1+m_2}\right)\,u_1+ \left(\dfrac{2\,m_2}{m_1+m_2}\right)\,u_2\) .......(7)
and \( v_2 = \left(\dfrac{2\,m_1}{m_1+m_2}\right)\,u_1+ \left(\dfrac{\,m_2-m_1}{m_1+m_2}\right)\,u_2\) ........(8)
then, from equation (7)
\(v_1 = u_1\)
and from equation (8),
\(\Rightarrow v_2 = 2\,u_1 - u_2\)
\(m_1 = m_2 =m\)
The from equitation (7) and (8)
\(\Rightarrow v_1 = u_2\)
and \( v_2 = u_1\)
Conclusion
If the masses of two bodies are same, then the velocity of both bodies will get inter changed after collision.
\(\Rightarrow \dfrac{1}{2} m_1\,u_1^2 + 0 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2\) .........(1) [kinetic energy is conserved]
\(\Rightarrow m\,u_1 = m_1\,v_1 \,cos\,\theta_1+m_2\,v_2 \,cos\,\theta_2\) .......(2) [along x- axis]
\(\Rightarrow 0+0 = m_1\,v_1 \,sin\,\theta_1-m_2\,v_2 \,sin\,\theta_2\) .......(3) [along y- axis]
A \(\dfrac{1}{2} m_1\,u^2 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2\)
B \(m\,v_1=m_1\,v_1 \,cos\,\theta_1+m_2\,v_2 \,cos\,\theta_2\)
C \(0=m_1\,v_1 \,sin\,\theta_1-m_2\,v_2 \,sin\,\theta_2\)
D \(\dfrac{1}{2} m_1\,v^2 = \dfrac{1}{2} m_1\,v_1^2\,cos^2\,\theta_1+\dfrac{1}{2} m_2\,v_2^2\,cos^2\,\theta_2\)