A stone is dropped from a height h. It hits the ground with a certain momentum p. If the stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by A) 68% B) 41% C) 200% D) 100% - Padhle.Online
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# A stone is dropped from a height h. It hits the ground with a certain momentum p. If the stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by A) 68% B) 41% C) 200% D) 100%

### Explanation

A stone is dropped from a height of h in the first scenario. We may deduce from the equations of motion that the velocity of the stone in free fall is given by:

v = √2gh——(1)

where

• g is the acceleration due to gravity
• h is the height from which the stone is dropped.

Also, the momentum of the stone is given by

p=mv——(2)

where

• p is the momentum of the stone
• m is the mass of the stone
• v is the velocity of the stone

Let this be equation 2. Substituting equation 2 in equation 1, we have

p = mv = m√2gh−−−(3)

Now, let us move on to the second case.

Here, the stone is dropped from a height 100% more than the previous height.

If we call this height H, it is given by

H = h + 100/100h = 2h

Again, if we take the velocity of this stone to be V, it is given by

V=√2gH =√2g(2h) 2√gh−−–(4)

Similarly, if the momentum of the stone in the second case is P, it is given by

P = mV = m2√gh−−–(5)

Now, to calculate the change in momentum, we subtract equation 3 from equation 5, as follows

P−p=mV−mv=m2√gh – m√gh = m√gh (√2-1) = 0.41m√gh

finally, to get the change in momentum in percentage, we take the ratio of this change in momentum to the original momentum and multiply by 100%.

This is shown as follows.

P−p/p x 100% 0.41m√gh/ m√gh x 100% =41%

Principal, Babu Daudayal SVM, Mathura