Q.Two different unknown masses A and B collide. A is initially at rest when B has a speed v. After the collision B has a speed v/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after the collision.
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Imagine two pool balls on a table. You hit one, it slides across the felt, and then it strikes the other ball at an angle — not dead centre, but off to the side. The balls don't just move along the same straight line after the hit; they scatter in different directions. That's a two-dimensional collision.
In one dimension, everything happens along a line — think of two trains bumping on a track. But in two dimensions, the collision is oblique: the objects approach each other at some angle, and after the collision they fly off in directions that are not along the original line of motion. The key insight is that momentum is a vector, and it is conserved component by component.
Note
A two-dimensional collision is any collision where the velocities before and after impact are not all along a single straight line. The motion happens in a plane.
The Core Idea: Momentum Conservation in Two Axes
Momentum is a vector — it has both magnitude and direction. When two objects collide, the total momentum vector before the collision equals the total momentum vector after the collision. That single vector equation breaks into two independent scalar equations, one for each perpendicular direction.
We choose two perpendicular axes — typically the x-axis and y-axis — that lie in the plane of motion. Then:
The total x-component of momentum before the collision equals the total x-component after.
The total y-component of momentum before the collision equals the total y-component after.
These two equations are completely independent. Nothing that happens along x affects the y-momentum conservation, and vice versa. This is the entire mathematical engine of two-dimensional collision analysis.
m1u1+m2u2=m1v1+m2v2
In components:
m1u1x+m2u2x=m1v1x+m2v2x
m1u1y+m2u2y=m1v1y+m2v2y
Here u are velocities before collision, v are velocities after, and m are masses.
What About Energy?
Momentum is always conserved in any collision (provided no external forces act). Energy is a separate story.
Elastic collision: Kinetic energy is also conserved. This gives you a third equation (scalar, not vector) that relates the speeds.
Inelastic collision: Kinetic energy is not conserved — some is lost to heat, sound, or deformation. You only have the two momentum equations.
Watch out
Do not assume energy conservation unless the problem explicitly says "elastic collision" or "perfectly elastic." Most real collisions are inelastic.
The Typical Problem Setup
Here is how a standard two-dimensional collision problem looks:
Object 1 (mass m1) moves with known velocity u1 along the x-axis.
Object 2 (mass m2) is initially at rest (u2=0).
They collide obliquely. After collision, object 1 moves at an angle θ1 to the x-axis with speed v1, and object 2 moves at an angle θ2 to the x-axis with speed v2.
You are typically asked to find some of these unknowns. The momentum equations give:
m1u1=m1v1cosθ1+m2v2cosθ2(x-component)
0=m1v1sinθ1−m2v2sinθ2(y-component)
The minus sign in the y-equation appears because the two objects usually scatter to opposite sides of the x-axis — one with positive y, the other with negative y.
Tip
Always draw a clear diagram showing the velocities before and after, with angles measured from a chosen reference axis. Label everything. This single step prevents most sign errors.
Why Two Axes Are Enough
You might wonder: why only x and y? Because the collision happens in a plane — two dimensions. Any vector in that plane can be fully described by its components along two perpendicular axes. There is no third independent direction. So two equations from momentum conservation, plus possibly one from energy conservation, give you up to three equations to solve for unknowns.
Step 1. Set up axes with B's original direction of motion along the x-axis. Before collision: A is at rest (momentum =0); B has momentum mBv along +x.
Step 2. After collision, B moves with speed v/2 perpendicular to its original direction -- take this as the +y direction, so B's final momentum is mB(v/2)j^, with zerox-component remaining in B.
Step 3. Momentum conservation along x:mBv=mAvA,x+0⇒vA,x=mAmBv.
Step 4. Momentum conservation along y:0=mAvA,y+mB(v/2)⇒vA,y=−2mAmBv (negative, i.e. opposite to B's deflection).
Step 5. The angle A makes with the x-axis (B's original direction): tanθ=vA,xvA,y=(mB/mA)v(mB/2mA)v=21 -- notice the unknown mass ratio mB/mAcancels out completely, so the angle is fixed regardless of what the (unknown) masses actually are. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2022Set sz1 mark
Q.In which collision, the colliding bodies move at certain angles before and after collision?
›Reveal solutionSolution
A collision where the velocities of the bodies (before and/or after impact) are not along a single common line is called an oblique or two-dimensional collision.
Collisions are classified by the geometry of the velocities involved:
A head-on (direct) collision: both bodies move along the same straight line (the line joining their centres) before and after collision -- this is a one-dimensional collision.
An oblique collision: the bodies approach and/or separate at some angle to each other rather than along a single line -- this is a two-dimensional (or oblique) collision. Momentum is still conserved, but must be resolved and conserved separately along two perpendicular directions (e.g. x and y components).
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