Question 1

In a scattering experiment, a particle of mass 2 m collides with another particle of mass m, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation o

Accepted Answer

2 m v_1=2 m v_ 1 f +2 m v_ 2 f ....(i) 2 m m_ 1 f =m v_ 2 f ....(ii) 1 2 (2 m) v_1^2+ 1 2 m(0)^2= 1 2 (2 m) v_ 1 f ^2+ 1 2 m v_ 2 f ^2 2 v_1^2=2 v_ 1 f ^2+v_ 2 f ^2 ....(iii) From (i), (ii), (iii), aligned & 3 v_ 1 f ^2-4 v_1 v_ 1 f +v_1^2=0 \ & (-4 v_1 )^2-4(3) (v_1^2 ) 0 \ & ^2 3 4 \ & ^2 3 2 \ & = 6 aligned

Question 2

Paragraph II: Two particles, 1 and 2 , each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their

Accepted Answer

At t _0=0 Before collision After collision aligned v _ CM & = m a 2 + m a m + m \ v _ CM & = 3 a 4 \ V _ CM a & = 3 4 \ V _ CM a & =0.75 aligned

Question 3

Two particles, 1 and 2 , each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass

Accepted Answer

t_0= 2 = T 4 Particles are at extreme position After collision in C-frame using WET, aligned & W _ spring = K \ & 1 2 k (2 ~b )^2- 1 2 k (2 a )^2=2 1 2 ~m ( a 4 )^2 ( k = spring constant ) \ & 4 ~kb ^2-4 ka ^2=2 m a ^2 16 2 k m \ & 4 ~b ^2= 17 4 a ^2 \ & 4 ~b ^2 a ^2 =4.25 aligned

Question 4

A bar of mass M = 1 . 00 &#160; kg and length L = 0 . 20 &#160; m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass

Accepted Answer

Before: After: Applying conservation of angular momentum about point O , we get m v 0 L 2 = M L 2 3 &#969; - m v L 2 &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . i As the collision is elastic,  e = 1 &#8658; velocity&#160;of&#160;separation(after&#160;collision) velocity&#160;of&#160;approach(before&#160;collision) = 1 &#8658; L &#969; 2 - - v v 0 = 1 &#8658; v = v 0 - L &#969; 2 &#160; &#160; &#160; &#160; &#160; &#160; . . . ii Using equation i  and ii , we can write m v 0

Center of Mass Momentum and Collision — Previous Year Questions & Solutions