Question

A stone is dropped ftom a height h.Its hit the ground with certain momentum

Answer 1

Explanation:

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.

Answer 2

Answer:

Hint: Momentum of a body under free fall refers to the property of the body which determines how much effort is required to stop the body under free fall. Momentum of a body is equal to the product of mass of the body and velocity of the body. Change in momentum refers to the difference between the initial momentum and the final momentum.

Complete step by step answer:

In the first case, a stone is dropped from a height h. From the equations of motion, we can derive that the velocity of the stone under free fall is given by

v=2gh−−−√

where

gis the acceleration due to gravity and h is the height from which the stone is dropped.

Let this be equation 1.

Also, the momentum of the stone is given by

p=mv

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=m2gh−−−√

Let this be equation 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+100100h=2h

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

V=2gH−−−−√=2g(2h)−−−−−√=2gh−−√

Let this be equation 4.

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

P=mV=m2gh−−√

Let this be equation 5.

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

P−p=mV−mv=m2gh−−√−mgh−−√=mgh−−√(2–√−1)=0.41mgh−−√

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−pp×100%=0.41mgh−−√mgh−−√×100%=41%