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Chapter 9: Problem 70

An object collides elastically with an equal-mass object initially at rest. If the collision isn't head-on, show that the final velocity vectors are perpendicular.

Short Answer

Expert verified
Based on the conservation of momentum and kinetic energy, and the given conditions, it can be shown that for an elastic collision where one of the equal-mass objects is initially at rest, the final velocity vectors of the two objects are perpendicular post collision. This was shown mathematically by calculating the angle between the vectors, which turned out to be 90 degrees.

Step by step solution

01

Establishing the conservation of momentum

Establish the law of conservation of momentum for this scenario. If the initial velocities (pre-collision) of Ball 1 and Ball 2 are \(v_1\) and \(v_2\) respectively, and the final velocities (post-collision) of Ball 1 and Ball 2 are \(u_1\) and \(u_2\) respectively, the conservation of momentum can be written as: \[m \cdot v_1 + m \cdot v_2 = m \cdot u_1 + m \cdot u_2\]where \(m\) is the mass of each object (equal for both objects as per problem statement).
02

Establishing the conservation of kinetic energy

Establish the conservation of kinetic energy for this scenario. Using the previous notation, conservation of kinetic energy can be written as: \[\frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2\] where \(m\) is the mass of each object (equal for both).
03

Implementing the given conditions

Taking into account the conditions mentioned: Ball 2 was initially at rest, so \(v_2 = 0\), and the collision is elastic. Thus, the equations for conservation of momentum and kinetic energy are simplified to:\[m v_1 = m u_1 + m u_2 \tag{1}\]\[\frac{1}{2} m v_1^2 = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 \tag{2}\]
04

Showing that the velocity vectors are perpendicular post-collision

Square equation (1), we get \[m^2 v_1^2 = m^2 (u_1^2 + u_2^2 + 2u_1 u_2 cosα)\]where α is the angle between \(u_1\) and \(u_2\).Next, compare this with equation (2), we find \(2u_1 u_2 cosα = 0\). Since \(u_1\) and \(u_2\) are non-zero (Ball 1 gave motion to Ball 2, and it also moved after collision), it implies that \(cosα = 0\). Hence, α = 90 degrees. This shows that the velocity vectors are perpendicular after the collision.

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