Explanation Given: The weight of the marble is W = 0.2 lb . The initial velocity of the marble at A is v A = 75 ft/s . The coefficient of the restitution between the marble and the wall is e = 0.5 . Draw the free body diagram of projectile motion of the marble as shown in Figure (1). Write the formula for position of the marble along x − a x i s . s x = ( s 0 ) x + ( v 0 ) x t (I) Here, s x is the final position of the marble along x − a x i s , ( s 0 ) x is the initial position of the marble along x − a x i s , ( v o ) x is the initial velocity of the marble along x − a x i s and t is the time. Write the formula for position of the marble along y − a x i s . s y = ( s 0 ) y + ( v 0 ) y t + 1 2 g t 2 (II) Here, s y is the final position of the marble along y − a x i s , ( s 0 ) y is the initial position of the marble along y − a x i s , ( v o ) y is the initial velocity of the marble along y − a x i s , g is the acceleration due to gravity and t is the time. Write the formula for velocity of the marble along y − a x i s . v y = ( v 0 ) y + g t (III) Here, v y is the final velocity of the marble along y − a x i s , ( v 0 ) y is the initial velocity of the marble along y − a x i s , g is the acceleration due to gravity and t is the time Write the principle of conservation of linear momentum. ∑ m v 1 = ∑ m v 2 (IV) Here, m is the mass, v is the velocity and the suffixes 1 , 2 indicates the initial and final stages. Here, the concrete wall does not move during the impact. Hence the velocity along x ' − a x i s is conserved. Write the formula for coefficient of restitution ( e ) along y ' − a x i s . e = 0 − ( v B ) y ' 2 ( v B ) y ' 1 − 0 (V) Here, v B is the velocity of the marble at B , the suffix y ' indicates the velocity is along y ' − a x i s and suffixes 1 , 2 indicates the before and after impact respectively. Write the formula for direction ( θ ) of the marble. From Figure (1). tan θ = ( v ) y ( v ) x (VI) Conclusion: Refer Figure (1a). The initial velocity of the marble along x − a x i s is ( v 0 ) x = v A cos 45 ° = ( 75 ft/s ) cos 45 ° = 53.033 ft/s The initial velocity of the marble along y − a x i s is ( v 0 ) y = v A sin 45 ° = ( 75 ft/s ) sin 45 ° = 53.033 m/s The final position of the marble along x − a x i s is s x = 100 ft . Calculate the time required to strike the wall at B . Substitute 100 ft for s x , 0 for ( s 0 ) x and 53.033 ft/s for ( v 0 ) x in Equation (I). 100 ft = 0 + ( 53.033 ft/s ) t t = 100 53.033 t = 1.886 s The marble requires the time of t = 1.886 s to strike the wall at B . Calculate the velocity of the marble along y − a x i s at the time t = 1.886 s . Substitute ( v B ) y 1 for v y , 53.033 ft/s for ( v 0 ) y , − 32.2 ft/s 2 for g and 1.886 s for t in Equation (III). ( v B ) y 1 = 53.033 ft/s + ( − 32.2 ft/s 2 ) ( 1.886 s ) = − 7.6962 ft/s Here, the negative sign indicates, the velocity is acting downward. The velocity of the marble along x − a x i s at the time t = 1.886 s is ( v B ) x 1 = ( v 0 ) x = 53.033 ft/s Calculate the net velocity of the marble before striking the wall at B . ( v B ) 1 = ( v B ) x 1 2 + ( v B ) y 1 2 Substitute 53.033 ft/s for ( v B ) x 1 and − 7

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Author: HIBBELER

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Chapter 15, Problem 78P

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The speed of the marble after it rebounds from the wall.

Solution Summary: The author calculates the speed of the marble after it rebounds from the wall, and the velocity after striking the concrete wall.

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The speed of the marble after it rebounds from the wall.

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