Advertisement Remove all ads

Consider a Head-on Collision Between Two Particles of Masses M1 and M2. the Initial Speeds of the Particles Are U1 and U2 in the Same Direction. the Collision Starts at T = 0 - Physics

Sum

Consider a head-on collision between two particles of masses m1 and m2. The initial speeds of the particles are u1 and u2 in the same direction. the collision starts at t = 0 and the particles interact for a time interval ∆t. During the collision, the speed of the first particle varies as \[v(t) = u_1 + \frac{t}{∆ t}( v_1 - u_1 )\]
Find the speed of the second particle as a function of time during the collision. 

Advertisement Remove all ads

Solution

It is given that:
Speed of the first particle during collision, \[v(t) = u_1 + \frac{t}{∆ t}( v_1 - u_1 )\]v 
Let v' be the speed of the second particle, during collision.

On applying the law of conservation of linear momentum on both particles, we get:

m1u1 + m2u2 = m1v(t) + m2v'

\[\Rightarrow m_1 u_1 + m_2 u_2 = m_1 u_1 + m_1 \times \left( \frac{t}{∆ t} \right)( v_1 - u_1 ) + m_2 v'\]

\[\text{ On dividing both the sides by m} _2 , \text{ we get: }\]

\[ u_2 = \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 ) + v'\]

\[\Rightarrow v' = u_2 - \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 )\]
The speed of the second particle during collision can be written as a function of time and is given by the expression,\[u_2 - \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 )\] .

Concept: Momentum Conservation and Centre of Mass Motion
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 9 Centre of Mass, Linear Momentum, Collision
Q 33 | Page 162
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×