All Exams Test series for 1 year @ ₹349 only
Question

A person jumps from a height on soft sand. Which one of the following is the correct reason for less likely injury to the person?

This question was previously asked in
NDA 2 2026 GAT Question Paper (13-Sep-2026)
The correct answer is
The sand increases the impact time.

Soft Sand Injury Prevention Physics

When landing after a jump, the body experiences a change in momentum. The force experienced during this landing depends on how quickly this momentum changes. The principle that explains this is the impulse-momentum theorem.

Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse (\(J\)) applied to an object is equal to the change in its momentum (\(\Delta p\)). Impulse is calculated as the average force (\(F_{avg}\)) multiplied by the time interval (\(\Delta t\)) over which the force acts.

The formula is: \(J = \Delta p = F_{avg} \times \Delta t\)

For a person jumping from a certain height, the change in momentum (\(\Delta p\)) when they stop upon landing is constant, regardless of the surface. Therefore, to minimize the injury-causing force (\(F_{avg}\)), the impact time (\(\Delta t\)) must be increased.

Role of Soft Sand

Soft sand provides a cushioning effect. When a person lands on it, the sand deforms, allowing the person's body to decelerate over a longer period. This means the impact time (\(\Delta t\)) is increased.

According to the impulse-momentum equation (\(F_{avg} = \frac{\Delta p}{\Delta t}\)), if \(\Delta t\) increases while \(\Delta p\) remains constant, the average force (\(F_{avg}\)) experienced by the person decreases. This reduction in force makes the landing less likely to cause injury.

Evaluating Other Options

  • Option 1: The sand exerts a reaction force (Newton's Third Law).
  • Option 2: The sand increases the impact time, which in turn reduces the force, but it doesn't primarily reduce the change in momentum itself.
  • Option 4: The sand does not increase momentum; it facilitates a decrease in momentum over a longer time, thus reducing the peak force.

Therefore, the increase in impact time is the correct reason for less likely injury.

Was this answer helpful?

Similar Questions

  1. The impulse on a particle due to a force acting on it during a given time interval is equal to the change in its

  2. A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?
  3. Two astronauts X and Y float in deep space, far from external forces. Initially, they are at rest relative to one another. X throws a tool towards Y. Consider the two astronauts and the tool as one system. Which one of the following statements is correct?


Important Questions from Conservation of Linear Momentum

  1. A heavy particle of rest mass M while moving along the positive z - direction, decays into two identical light particles with rest mass m (where M > 2m). The maximum value of the momentum that any one of the lighter particles can have in a direction perpendicular to the z-direction, is

  2. A metal ball with the momentum mv strikes a wall and bounces back. The change in the ball's momentum is ideally

  3. A mass of 4 kg moving at 3 m/s collides with a mass of 6 kg moving at 2 m/s in the opposite direction and they stick together. The combined mass has velocity of:

  4. In an orbital motion, the angular momentum vector is:

  5. Body A of mass $m$ moving with speed $u$ collides with another body B of mass $3m$, at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is:
Need Expert Advice?
Test Series
NDA img
Defence
NDA 2026 Mock Test Series (Latest Pattern)
503 Tests 1 Tests Free
1066 Attempts
4.6(137)
English, Hindi

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App