PHysics Question?

Question

A 0.100-kg ball is thrown straight up into the air w/ an initial speed of 15.0 m/s. Find the momentum of the ball (a) at its maximum height and (b) halfway up to its maximum height.

Answer 1

a) is easy. At the ball's maximum height, it stops for a moment before it begins to fall. It has a velocity of zero, and momentum is the product of velocity and mass, so the momentum is also zero.

b) is more difficult. We assume the "maximum height" refers to maximum height above the thrown height, and we can treat the height from which it was thrown as if it were h = 0.

I'd use energy balance to determine the maximum height. If KE is kinetic energy and GPE is gravitational potential energy, you have KE + GPE being constant due to conservation of energy, so KE_0 + GPE_0 = KE_f + GPE_f, where the subscript 0 refers to initial conditions and f refers to final conditions. Remember, KE = 0.5*mv^2 and GPE = mgh. So you get 0.5*m*(v_0)^2 + mg(h_0) = 0.5*m*(v_f)^2 + mg(h_f). h_0 and v_f are zero, and m cancels out, so you're left with 0.5*(v_0)^2 = mg(h_f), where v_0 was given, so h_f is the only unknown.

Once you solve for h_f, repeat the energy balance using the intial conditions and a height of (h_f)/2, where the only unknown would be the velocity at this height. Solve for the velocity and calculate the momentum, which is still just the product of the mass and velocity (which you just found).

Answer 2

DavidK93 is right on the first part. At its peak, it isn't moving so velocity=0m/s and so its momentum is also 0 kgxm/s at the peak.
Halfway, up, though......let's see.
Using work-energy principle is easier than standard kinematics on this problem. So it's KE at the start is
KE=(0.5)(0.1)(15)(15) = 11.25 Joules
So its PE will also be 11.25J at the peak. Its peak height will be PE = mgh
11.25J = (0.1)(9.81)(h)
height = 11.47meters
half of the height would be 5.73meters
and it will have half of the K.E. there, because it's gained half of the PE it will gain.
so the KE it will possess there will be 11.25 / 2 or 5.625J
We still need velocity
KE = (0.5)(m)(v)(v)
5.625 J = (0.5)(0.1)(v)(v)
velocity at halfway up is....
10.6 m/s
so momentum = (mass)(velocity)
momentum = (0.1kg)(10.6m/s) = 1.06 kgxm/s

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Answer 3

a) at maximum height the velocity is zero. The momentum is zero

b) half way to the maximum height
max H = v^2/2g = 15^2/19.62 = 11.468 meters
H/2 = 5.734 meters
v^2 = 15^2 - 19.62(5.734) = 112.50
v=sqr(112.50) = 10.606 m /s
momenturm = mv = 0.10(10.606) = 1.06 Kg.m/se

Answer 4

At maximum height it's speed is zero, at the half way point it's speed is 8.3m/s.