Explanation Given: The mass of the ball is m = 300 g . The velocity of the ball at A is v A = 25 m/s . The coefficient of the restitution between the ball and the ground surface is e = 0.4 . Draw the free body diagram of the ball’s projectile motion as shown in Figure (1). Refer Figure (1). By symmetrical property of the trajectory v A = v B = 25 m/s . Refer figure (1). Here, no impulsive force is acting along the x − a x i s . Hence the linear momentum is conserved along the x − a x i s . Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 m ( v B ) x = m ( v ' B ) x (I) Here, m is the mass, v B is the velocity of the ball at B and the suffixes x indicates the along x − a x i s . When, the ground the surface cannot move during the impact. Write the formula for coefficient of restitution ( e ) . e = 0 − ( v ' B ) y ( v B ) y − 0 (II) Here, v B , v ' B is the velocity of the ball at B and the suffixes y indicates the along y − a x i s . Write the formula for direction ( θ ) of the ball. From Figure (1). tan θ = ( v ' B ) y ( v ' B ) x (III) Conclusion: Refer Figure (1). Substitute 300 g for m and 25 ( cos 30 ° ) m/s for ( v B ) x in Equation (I). ( 300 g × 1 kg 1000 g ) × 25 ( cos 30 ° ) m/s = ( 300 g × 1 kg 1000 g ) ( v ' B ) x 6.4952 = 0

15th Edition

ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 69P

To determine

The magnitude and the direction (θ)

of the velocity of the ball:

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Chapter 15, Problem 68P

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