Question

A person of mass jumps from a height 50m and lands on the ground in two ways: i) he flexes his knees and brought to rest in 1s. ii) he does not flex his knees and rest in 0.1s. calculate the force in both the cases and in which case less damage is done to the body

Answer 1

SOLUTION.

As we know that a body of some mass when dropped from certain height attain velocity due to gravity so in order to stop the body we need to apply external force. Which is proportional to the velocity of the body. so it is clear that a body travelling with great speed need great force to stop it's motion. From this I want to conclude that force is directly proportional to momentum of the body.

But what about time taken to stop the body hence if higher magnitude of speed stop with in seconds than you can say that magnitude of force is inversely proportional to the time taken to stop the body.

From both of these discussion we get

F = dp/dt

where dp is the change in momentum.

dt is the time taken to change momentum.

Height = 50m

Using eq of motion

v² = u² + 2gh

v² = 0² + 2×10×50

v² = 1000

v = 10√10m/sec

We know that momentum can be written as...

P = mv

P = 10m√10.

In both the cases initial momentum is P

Final momentum = 0

∆p (Change in momentum) = 0 - 10m√10

∆p = -10m√10

• Part 1

Time taken(dt) = 1sec

Force = dp/dt

Force = -10m√10 N

• Part 2

Time taken(dt) = 0.1

Force = dp/dt

Force = -100m√10 N

This negative sign indicate that Force applied opposite to that of momentum.

IN SECOND CASE BODY FACE MORE DAMAGE BECAUSE THIS FORCE DECREASE SAME MOMENTUM BUT IN LESSER TIME.

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