Question

1. a 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball. as a result, the incident ball is deflected 30.0° from its initial path. (a) draw a vector diagram showing the momentum of both balls after the collision. (b) what is the velocity of the incident ball after the collision? (c) what is the velocity of the struck ball after the collision?

Explanation:

Step1: Recall elastic - collision principles

In an elastic collision of two identical balls where one is initially stationary, the momentum and kinetic - energy are conserved. Let the mass of each ball be \(m\), the initial velocity of the incident ball be \(v_0 = 0.40\ m/s\), and the initial velocity of the struck ball be \(u_0=0\).

Step2: Analyze momentum conservation in two - dimensions

For a two - dimensional elastic collision, if the incident ball is deflected by an angle \(\theta = 30.0^{\circ}\) from its initial path. Let the velocity of the incident ball after the collision be \(v_1\) and the velocity of the struck ball be \(v_2\).
By conservation of momentum in the x - direction (initial direction of the incident ball): \(mv_0=mv_1\cos\theta+mv_2\cos\varphi\), and in the y - direction: \(0 = mv_1\sin\theta - mv_2\sin\varphi\). Since the collision is elastic, \(v_0 = v_1 + v_2\) (from kinetic - energy conservation for identical masses \(m_1 = m_2=m\): \(\frac{1}{2}mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\), which simplifies to \(v_0^2=v_1^2 + v_2^2\) for \(m_1 = m_2\)). Also, from the y - momentum equation \(v_1\sin\theta=v_2\sin\varphi\), and from the x - momentum and kinetic - energy equations for identical masses in elastic collision, we know that the two balls move at right - angles to each other after the collision (\(\theta+\varphi = 90^{\circ}\)).

Step3: Solve for velocities

Since \(\theta = 30^{\circ}\) and \(v_0 = v_1 + v_2\) and \(v_0^2=v_1^2 + v_2^2\), we can use trigonometry. If we consider the right - triangle formed by the velocities, we know that \(v_1 = v_0\cos\theta\) and \(v_2 = v_0\sin\theta\).

(a)

The vector diagram will have the initial momentum vector of the incident ball as a horizontal arrow. After the collision, the incident ball's momentum vector will be at an angle \(\theta = 30^{\circ}\) to the horizontal with magnitude \(mv_1\), and the struck ball's momentum vector will be at an angle \(\varphi = 60^{\circ}\) to the horizontal with magnitude \(mv_2\). The sum of the two momentum vectors after the collision will equal the initial momentum vector of the incident ball.

(b)

The velocity of the incident ball after the collision: \(v_1=v_0\cos\theta\). Substituting \(v_0 = 0.40\ m/s\) and \(\theta = 30^{\circ}\), we get \(v_1 = 0.40\times\cos30^{\circ}=0.40\times\frac{\sqrt{3}}{2}\approx0.346\ m/s\).

(c)

The velocity of the struck ball after the collision: \(v_2=v_0\sin\theta\). Substituting \(v_0 = 0.40\ m/s\) and \(\theta = 30^{\circ}\), we get \(v_2 = 0.40\times\sin30^{\circ}=0.20\ m/s\).

Answer:

(a) Draw a horizontal arrow for the initial momentum of the incident ball. After the collision, draw an arrow for the incident - ball's momentum at a \(30^{\circ}\) angle to the horizontal and an arrow for the struck - ball's momentum at a \(60^{\circ}\) angle to the horizontal such that the sum of the two post - collision momentum vectors equals the initial momentum vector.
(b) \(v_1\approx0.346\ m/s\)
(c) \(v_2 = 0.20\ m/s\)