Question

A baseball of mass m1 = 0.45 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.35 m. The second ball m2 = 0.61 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 4.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity.) What is the angle that the string makes with the vertical at the highest point of travel in degrees?

Answer 1

Answer:

0.947 rad or 54.27 degrees

Explanation:

Suppose the collision is elastic, meaning the momentum is preserved.

Before the collision $v_2 = 0 m/s$ and the second ball has no momentum

Right after the collision $v_1 = 0 m/s$ and the first bast ball has no momentum.

Therefore momentum of the first baseball has been transferred to the 2nd ball

$m_1v_1 = m_2v_2$

$4.5 * 0.45 = 0.61v_2$

$v_2 = \frac{4.5 * 0.45}{0.61} = 3.32 m/s$

After the collision, the second ball would gained kinetic energy, which then would be transferred to potential energy once it reaches its highest point:

By the law of energy conservation:

By law of energy conservation:

$E_k = E_p$

$\frac{mv^2}{2} = mgy$

$v^2 = 2gy$

$y = \frac{v^2}{2g} = \frac{3.32^2}{2*9.81} = 0.562 m$

So at its highest point, the ball is 0.562 m from the lowest point. Since the ball is hanging on a 1.35 m string, we can calculate the vertical distance from there to the swinging point:

1.35 - 0.562 = 0.788 m

Finally, the angle that string makes with the vertical at the highest point is

$\alpha = cos^{-1}(\frac{0.788}{1.35}) = 0.947 rad = 54.27^o$

Answer 2

The angle that the string makes with the vertical at the highest point of travel is approximately 54 degrees.

Given the data in the question;

• Mass of first baseball; $m_1 = 0.45kg$

• Mass of second baseball; $m_2 = 0.61kg$

• Initial horizontal velocity; $v_1 = 4.5 m/s$

• Length of string; $l = 1.35m$

First, we determine final velocity $v_2$

From Conservation of Momentum:

$m_1v_1 = m_2v_2$v

We make $v_2$, the subject of the formula

$v_2 = \frac{m_1v_1}{m_2}$

We substitute our given values into the equation

$v_2 = \frac{0.45kg\ *\ 4.5m/s}{0.61kg}\\\\v_2 = \frac{2.025kg.m/s}{0.61kg}\\\\v_2 = 3.3m/s$

Now, we find the maximum height $h_m$ reached

At maximum height, the kinetic energy of the second ball is transformed into potential energy.

Using Mechanical Energy Conservation:

Kinetic Energy = Potential Energy

$\frac{1}{2}mv^2 = mgh$

We make $h$, the subject of the formula

$h = \frac{v_2^2}{2g}$

We know that $g = 9.8m/s^2$, so we substitute our values into the equation

$h = \frac{(3.3m/s)^2}{2\ *\ 9.8m/s^}$

$h = \frac{10.89m^2/s^2}{19.6m/s^2} \\\\h = 0.5556m$

So, maximum height $h_m = 0.5556m$

Now, we calculate the Angle that the string makes;

$\theta = cos^{-1}( [l - hm] / l )$

We substitute in our values

$\theta = cos^{-1}( [1.35m - 0.5556m] / 1.35m )\\\\\theta = cos^{-1}( 0.7944m/ 1.35m )\\\\\theta = cos^{-1}( 0.5884)\\\\\theta = 53.956^o\\\\\theta = 54^o$

Therefore, the angle that the string makes with the vertical at the highest point of travel is approximately 54 degrees.

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