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Question

A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground is changed by ________?


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Solution

Step 1: Given Data

Given: Momentum =p;

Let the mass of the ball bem.

The height is incremented by 100% with respect to the previous height:

If the previous height was ‘h’, the new height H becomes ‘2h’.

Step 2: Case 1: When the ball is dropped from a height ‘h

Initial velocity (u1) =0

Let the final velocity be v1

Net momentum =p=mv1

Step 3: Case 2: When the ball is dropped from a height H,

Initial velocity (u2)=0

Let the final velocity be v2

Net momentum =p'=mv2(i)

Step 4: Establish the Equation of Proportionality

From the third equation of uniformly accelerated motion:

v2=u2+2as

v22=u22+2gHv22=(0)2+2gHv22=2gHv2=2gH(equation(ii))

Replacing the v2by 2gHin equation (i)

We get: p'=m2gH

We can derive from the equation that p'H

In general, we can say that: ph

Step 5: Finding the relationship between momentum in case 1 and case 2:

p'p=Hhp'p=2hhp'p=2=1.414p'=1.414p

Step 6: Calculating the percentage change in the momentum:

Percentage change=(p'-p)p×100=(1.414p-p)p×100=41.4%

Therefore, the momentum when it hits the ground is changed by 41.4%.


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