Explanation Given information: The velocity of the ball A ( v A ) is v 0 . The angle ( θ ) is 45 ° . The coefficient of restitution between each car ( e ) is 0.8. Assume no friction. Calculation: Show the momentum diagram of the ball A and ball B as Figure (1). Consider the ball A in tangent ( t ) direction. The expression for the conservation of momentum equation for the ball A as follows: m A v A sin θ = m A v ′ A t Here, v ′ A t is the velocity of the ball A in tangent ( t ) direction after impact and m A is the mass of the ball A . Substitute v 0 for v A . m A v 0 sin θ = m A v ′ A t v 0 sin θ = v ′ A t Substitute 45 ° for θ . v 0 sin 45 = v ′ A t v ′ A t = 0.707 v 0 Consider the ball B in tangent ( t ) direction. The velocity of the ball B ( v B ) will be zero as it rest initially. The expression for the conservation of momentum equation for the ball B as follows: m B v B = m B v ′ B t Here, v ′ B t is the velocity of the ball B in tangent ( t ) direction after impact and m B is the mass of the ball B . Substitute 0 for v B . m B ( 0 ) = m B v ′ B t 0 = v ′ B t Consider both the ball A and ball B in n t h direction. The expression for the conservation of momentum equation for both ball A and ball B as follows: m A v A cos θ + m B v B = m A v ′ A n + m B v ′ B n Here, v A is the initial velocity of ball A and v B is the initial velocity of ball B . Substitute m for m A , m for m B , v 0 for v A and 0 for v B . m v 0 cos θ + m ( 0 ) = m v ′ A n + m v ′ B n v 0 cos θ = v ′ A n + v ′ B n (1) The expression for the initial velocity of the ball A in n t h direction ( v A n ) as follows: v A n = v 0 cos θ The initial velocity of the ball B in n t h direction ( v B n ) is zero. The expression for the coefficient of restitution for the ball A and ball B as follows: v ′ B n − v ′ A n = e ( v A n − v B n ) Here, v A n is the initial velocity of the ball A in n t h direction and v B n is the initial velocity of the ball B in n t h direction. Substitute v 0 cos θ for v A n and 0 for v B n . v ′ B n − v ′ A n = e ( v 0 cos θ − 0 ) v ′ B n − v ′ A n = e v 0 cos θ (2) Solve the equation (1) and (2). 2 v ′ B n = v 0 cos θ + e v 0 cos θ 2 v ′ B n = v 0 cos θ ( 1 + e ) v ′ B n = v 0 cos θ ( 1 + e ) 2 Substitute 0

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 13.4, Problem 13.164P

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Find the velocity of ball A (v′¯A) and ball B (v′¯A) after impact.

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Chapter 13.4, Problem 13.163P

Chapter 13.4, Problem 13.165P

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Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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Find the velocity of ball A ( v ′ ¯ A ) and ball B ( v ′ ¯ A ) after impact.

Solution Summary: The author shows the momentum diagram of the ball A and ball B as Figure (1).

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Find the velocity of ball A (v′¯A) and ball B (v′¯A) after impact.

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