Easy Physics Question

[Righty123]

A 2.5kg stone is dropped from a height of 4m. What is the momentum on impact? (Ignore air resistance.)

After looking at the answer, the book utilized one of the big five to solve the question. The way I solved it, however, was that I got the PE (2.5x4x10) and subsequently got the v since at impact PE=0, and it is all converted to KE. From there, I just plugged the v into the mv (momentum) equation and got a different, incorrect, answer.

---

Members don't see this ad.

 

[DrChandy]

A 2.5kg stone is dropped from a height of 4m. What is the momentum on impact? (Ignore air resistance.)

After looking at the answer, the book utilized one of the big five to solve the question. The way I solved it, however, was that I got the PE (2.5x4x10) and subsequently got the v since at impact PE=0, and it is all converted to KE. From there, I just plugged the v into the mv (momentum) equation and got a different, incorrect, answer.

Use V(f)^2=V(i)^2+(xf-xi)2a

Using the approximation of g=a=10m/(s^2),
and substituting the values of (xf-xi)=4m, and v(i)=0
Yields a V(f) of ~9 m/s

The way you solved it also works:
mgh=1/2m(v)^2
w/ constant mass, yields gh=1/2m(v)^2
solve for v to get v=~9m/s

Now solve for P=mv=(2.5kg)x(9m/s)=22.5 (kgm)/s final momentum

 

[SpyPuts]

However you choose to work the problem, the answer should be:

p=m*v
p=m*[(2gh)^1/2]

If you notice, both methods of working the problem will converge to give a value of root(2gh) for the final velocity on impact.

From the linear motion equation:
(Vf)^2 = (Vi)^2 + 2a(Xf-Xi)
(Vf)^2 = 0 + 2gh
(Vf)^2 = 2gh
Vf = root(2gh)

Since there are no non-conservative forces acting:
Ui + Ki = Uf + Kf
mgh + 0 = 0 + (1/2)(m)(Vf)^2
(Vf)^2 = 2gh
Vf = root(2gh)

Hope that makes sense. Exponents always look terrible in typed text when there aren't formatting options available.