Question

A 4.00 −kg ball is dropped from a height of 14.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.50 kg and is 7.40 m in length. At the other end of the bar sits another 5.30 −kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision.

How high will the other ball go after the collision?

Answer 1

First, analyze the collision between the falling ball and the bar:

The falling bar has gravitational PE which is transformed into kinetic energy:

PE= KE

mgh = (1/2)mv²

Solving for v:

v= (2gh)^1/2 = (2 x 9.81 x 14. 0)^1/2 = 16.57 m/s. This is the speed at which the ball hits the bar.

Before the collision, the ball has linear momentum:

P=mv= 4 kg x 16.57 m/s = 66.28 kg m/s

If we study the ball - bar system, it's angular momentum (taking the pivot of the bar as reference) should be conserved:

L=r x P = 3.7 m x 66.28 kg m/s = 245.236 kg m²/s so we can find the angular speed that the bar acquires:

w= L / I,

where I is the moment of inertia of the bar with the two balls attached, relative to its center:

I= 1/12 (m_bar x l_bar²) + [1/2 m_ball1(l_bar/2)²] + [1/2 m_ball2(l_bar/2)²

I= 1/12 (5.5 x 7.40²) + 1/2 (4 x 3.7²) + 1/2(5.30 x 3.7²)

I= 88.75 kg m²

Then:

w= (245.236 kg m²/s) / 88.75 kg m² = 2.76 rad/s

So the other ball acquires the speed:

v₂= w (l_bar/2) = 2.76 rad/s x 3.7 m = 10.22 m/s

and the height can be found using energy balance for this other ball:

1/2 m₂v₂² = m₂gh₂

h₂= v₂² / 2g = (10.22 m/s)²/ (2 x 9.81 m/s²) = 5.34 m

The other ball would go 5.34 m after the collision.