Question

1. a 3.8 kg steel ball collides with a stationary 5.5 kg steel ball. as a result of this collision the two balls travel as shown. what was the original velocity of the 3.8 kg ball? v = 2.55 m/s v = 4.0 m/s m = 5.5 kg m = 3.8 kg

Explanation:

Step1: Apply conservation of momentum in x - direction

Initially, only the \(3.8\) kg ball has momentum in the x - direction (assuming the initial motion is along a line). After collision, we have components of momentum for both balls.
The initial momentum in x - direction \(P_{ix}=m_1v_{1i}\) (where \(m_1 = 3.8\) kg and \(v_{1i}\) is the initial velocity of \(3.8\) kg ball). After collision, \(P_{fx}=m_2v_2\sin32^{\circ}-m_1v_1\sin19^{\circ}\) (using \(\sin\) for x - components of velocities). By conservation of momentum \(P_{ix} = P_{fx}\), so \(m_1v_{1i}=m_2v_2\sin32^{\circ}-m_1v_1\sin19^{\circ}\).

Step2: Apply conservation of momentum in y - direction

Initially, momentum in y - direction \(P_{iy} = 0\). After collision, \(P_{fy}=m_2v_2\cos32^{\circ}+m_1v_1\cos19^{\circ}\). By conservation of momentum \(P_{iy}=P_{fy}\), so \(0 = m_2v_2\cos32^{\circ}+m_1v_1\cos19^{\circ}\) (this step is actually not needed as we can use the fact that in the absence of external forces, momentum is conserved in each direction. But for the problem, we focus on x - direction as initial motion is along a line (implied from the problem setup of finding initial velocity).
Substitute \(m_1 = 3.8\) kg, \(m_2 = 5.5\) kg, \(v_1=2.55\) m/s, \(v_2 = 4.0\) m/s into \(m_1v_{1i}=m_2v_2\sin32^{\circ}-m_1v_1\sin19^{\circ}\)

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Answer:

The original velocity of the \(3.8\) kg ball is approximately \(2.23\space m/s\)