(a) Explanation Given information: The weight of the curling stone ( W ) is 40 lb .. The diameter of the stone ( d ) is 11 in . . The width of the middle of stone A to the middle of stone B ( b ) is 5 in . . The velocity of stone A ( v A ) is 0 .5m/s . The velocity of stone B is at rest that is 0. The coefficient of restitution between each car ( e ) is 0.7. Calculation: Calculate the angle ( θ ) using the relation: cos θ = b d Substitute 5 in . for b and 11 in . for d . cos θ = 5 11 θ = cos − 1 ( 5 11 ) θ = 62.964 ° Show the impulse momentum diagram for the stone A as Figure (1). Show the impulse momentum diagram for the stone B as Figure (2). Take the mass of the stone A ( m A ) and mass of the stone B ( m B ) are equal. Consider the stone A in tangent ( t ) direction. The expression for the conservation of momentum equation for the stone A as follows: m A v A cos θ = m A v ′ A t Here, v ′ A t is the velocity of the stone A in tangent ( t ) direction after impact and m A is the mass of the stone A . Substitute v 0 for v A and 62.964 ° for θ . m A ( v 0 ) cos 62.964 ° = m A v ′ A t v ′ A t = 0.4545 v 0 Consider the stone B in tangent ( t ) direction. The velocity of the stone B ( v B ) will be zero as it rest initially. The expression for the conservation of momentum equation for the stone B as follows; m B v B = m B v ′ B t Here, v ′ B t is the velocity of the stone B in tangent ( t ) direction after impact and m B is the mass of the stone B . Substitute 0 for v B . m B ( 0 ) = m B v ′ B t 0 = v ′ B t Consider both the stone A and stone B in n t h direction. The expression for the conservation of momentum equation for both stone A and stone B as follows: m A v A sin θ + m B v B = m A v ′ A n + m B v ′ B n Here, v A is the initial velocity of stone A and v B is the initial velocity of stone B . Substitute m for m A , m for m B , v 0 for v A and 0 for v B . m v 0 sin θ + m ( 0 ) = m v ′ A n + m v ′ B n v 0 sin θ = v ′ A n + v ′ B n Substitute 62.964 ° for θ . v 0 sin 62.964 = v ′ A n + v ′ B n v ′ A n + v ′ B n = 0.8907 v 0 (1) The expression for the initial velocity of the stone A in n t h direction ( v A n ) as follows: v A n = v 0 sin θ The initial velocity of the stone B in n t h direction ( v B n ) is zero. The expression for the coefficient of restitution for the stone A and stone 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 stone A in n t h direction and v B n is the initial velocity of the stone B in n t h direction (b) To determine Show that if e = 1 , the final velocities of the stones form a right angle for all values of b .

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

12th Edition

ISBN: 9781260916942

Author: BEER

Publisher: MCG

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Chapter 13.4, Problem 13.165P

(a)

To determine

Find the velocity of stone A (v′A) and velocity of stone B (v′B) in terms of (v0).

(b)

To determine

Show that if e=1, the final velocities of the stones form a right angle for all values of b.

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Chapter 13.4, Problem 13.164P

Chapter 13.4, Problem 13.166P

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VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

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Find the velocity of stone A ( v ′ A ) and velocity of stone B ( v ′ B ) in terms of ( v 0 ) .

Solution Summary: The author calculates the velocity of stone A and B using the relation: