Physics · Laws Of Motion · NEET
Impulse equals CHANGE in momentum, and momentum is a vector. Take the wall's outward direction as positive. Before: the ball moves toward the wall, so its momentum is -mv. After: it moves away, so momentum is +mv. Change = final - initial = (+mv) - (-mv) = 2mv. The two mv values add because the direction flipped. Only mv would apply if the ball simply stopped (final momentum 0).
No. Speed is a scalar and it is unchanged, but VELOCITY is a vector and it reversed. Impulse depends on the change in the velocity vector, not on speed. So a same-speed rebound gives a large impulse of 2mv, while it might feel like 'nothing changed'. This is the single most common NEET trap here.
Split the velocity into two parts: the component PERPENDICULAR (normal) to the wall and the component PARALLEL to the wall. The wall is smooth, so it only pushes normally. The perpendicular component reverses; the parallel component stays the same. So impulse = 2m x (perpendicular component). If the angle is measured from the wall, use v sin(theta). If measured from the normal, use v cos(theta).
Impulse = change in momentum, measured in kg m/s or N s, and it does NOT need the contact time. Force = impulse divided by contact time (F = change in p / change in t), measured in newtons. If the question gives you the contact time, they want the force. If not, they can only ask for impulse.
The force on the ball by the wall points along the normal, away from the wall (it pushes the ball back). By Newton's third law, the force on the wall by the ball points into the wall, along the normal. For an angled hit, the force is STILL normal to the wall, not along the incoming path — a classic NCERT result from the billiard-ball example.
A rigid ball of mass m strikes a rigid wall at 60 degrees and gets reflected without loss of speed, as shown in the figure. The value of the impulse imparted by the wall on the ball is:
A ball of mass 0.15 kg is dropped from a height of 10 m, strikes the ground and rebounds to the same height. The magnitude of the impulse imparted to the ball is nearly (g = 10 m/s^2):
A ball of mass 0.5 kg is dropped from a height of 40 m. It hits the ground and rebounds to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (g = 9.8 m/s^2):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
An impulsive force is a very large force that acts for a very short time and produces a finite change in momentum — for example, the wall on a ball during a bounce. We usually cannot measure the force and the tiny contact time separately, but their product (the impulse = change in momentum) is easy to find.
For a head-on bounce off a wall: Impulse = change in momentum = m x v_after - m x v_before. If the ball rebounds with the same speed, this is 2mv. For an angled hit off a smooth wall, only the normal component reverses, so impulse = 2m x v_normal (which is 2mv cos(theta) if theta is from the normal, or 2mv sin(theta) if theta is from the wall).
Average force = impulse / contact time = change in momentum / change in time. For example, if the impulse is 2mv and the contact time is t seconds, then F = 2mv / t. The direction of this force is along the normal to the wall.
If the ball rebounds to a lower height, the collision is not perfectly elastic — some kinetic energy is lost, so the rebound speed is smaller than the incoming speed. You then compute the down speed and up speed separately using v = sqrt(2gh) for each height, and add their magnitudes for the impulse since the directions are opposite.
For a smooth (frictionless) wall, no — the wall can only push along its normal, so the parallel component of velocity is unchanged. Only the perpendicular (normal) component reverses. This is why impulse for an angled hit uses only the normal component.
