Body A having a mass of 2 kg and body B having a mass of 3 kg moving towards each other with velocities 4 m/s and 2 m/s, respectively, will collide and ______ in an elastic collision.
move in opposite direction
Concept:
Collisions are generally classified into two types: elastic and inelastic collisions.
This classification is based on the law of conservation, and the laws of collision state:
Perfectly elastic collision
If the laws of momentum conservation and kinetic energy hold well during the collision.
That is, initial velocity = final velocity

Inelastic collision
If the law of momentum conservation holds well during the collision while kinetic energy does not.
However, inelastic collisions follow the law of conservation of linear momentum.
Coefficient of restitution (e)
\(e = \frac{{Relative\;velocity\;after\;collision}}{{Relative\;velocity\;before\;collision}} = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\)
Calculation:
Let us assume that,
Mass of body A, mA = 2 kg
Block A, uA = 4 m/s . velocity
Mass of body B, mB = 3 kg
Velocity of block B, uB = -2 m/s
Now, according to the law of conservation of momentum
mA uA + mB uB = mA vA + mB vB
2 x 4 - 3 x (-2) = 2 x vA + 3 x vB
2 x vA + 3 x vB = 2 -----(1)
Now, for an inelastic collision, the coefficient of restitution is always one and can be expressed as
\(e=1=\frac{{{v}_{B}}-{{v}_{A}}}{{{u}_{A}}-{{u}_{B}}}=\frac{{{v}_{B}}-{{v}_{A}}}{4+2}\Rightarrow {{v}_{B}}-{{v}_{A}}=6\) ------(2)
Thus, comparing both quadratic equations, we obtain the values of the velocities of bodies A and B as follows
\({{v}_{B}}=\frac{14}{5}m/s\)
And now substituting this value into eq (2), we get, \({{v}_{A}}=-\frac{16}{5}m/s\)
Thus, we can see that body A will be moving in the opposite direction compared to body B.
Therefore, in this case, body A and body B will move in opposite directions in an elastic collision.
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