Question

A 29.0 g ball is fired horizontally with initial speed v0 toward a 110 gball that is hanging motionless from a 1.00 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 110 gball swings out to a maximum angle θmax = 51.0 ∘.

What is v0?

Answer 1

Answer 2

Let's start by finding the velocity of the 110 g ball just after the collision. We can use conservation of momentum to do this:

m1v1 + m2v2 = m1v1' + m2v2'

where m1 and m2 are the masses of the balls, v1 and v2 are their initial velocities, and v1' and v2' are their velocities just after the collision. Since the 110 g ball is initially at rest, we can simplify the equation to:

m1v1 = m1v1' + m2v2'

Solving for v1', we get:

v1' = (m1v1 - m2v2') / m1

Now let's use conservation of energy to find the initial velocity v0. Since the collision is perfectly elastic, the total kinetic energy of the system is conserved:

1/2 m1v1^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2

Substituting in the expression we found for v1', we get:

1/2 m1v1^2 = 1/2 m1(m1v1 - m2v2')^2 / m1 + 1/2 m2v2'^2

Simplifying and solving for v0, we get:

v0 = sqrt(2m2v2'^2 / (m1 - m2))

Now we just need to find the velocity v2' of the 29.0 g ball just after the collision. We can use conservation of momentum again:

m1v1 + m2v2 = m1v1' + m2v2'

v2' = (m1v1 + m2v2 - m1v1') / m2

Substituting in the values we know, we get:

v2' = (0.029 kg * v0 - 0.110 kg * 0) / 0.110 kg

v2' = 0.264 v0

Finally, we can use trigonometry to find the height that the 110 g ball swings to:

h = L - L cos(θmax) = L(1 - cos(θmax))

where L is the length of the string. Substituting in the values we know, we get:

h = 1.00 m (1 - cos(51.0°)) = 0.755 m

Since the ball swings to a height equal to its initial height, we can equate the potential energy before and after the collision:

mgh = 1/2 m2v2'^2

Solving for v0, we get:

v0 = sqrt(2gh / (1 - m2/m1))

Substituting in the values we know, we get:

v0 = sqrt(2 * 9.81 m/s^2 * 0.755 m / (1 - 0.110 kg / 0.029 kg))

v0 = 8.86 m/s

Therefore, the initial speed of the 29.0 g ball should be 8.86 m/s.