Question

Wall

The 1.5 kg ball A is released from rest from a height. As it descend hits a solid wall. If the coefficient of restitution is 0.85 determine : rebound height  after the impact and angular velocity for a 30° angle respect the release position (the horizontal)

Answer 1

ball mess= 1.5 kg height h angle- 30 egnt-h ball offc ert of restitution e#0 85 wall wall wall a) b) Fig 1 Given Mass of the ball is 1.5 kg Height of the ball at rest from the point of impact is ho Coefficient of restitution is CR 0.835 Assumption The motion of ball is curvilinear, i.e. the horizontal distance traveled is same as the vertical distance which is the height of ball at at rest from the point of impact To find Rebound Height of the ball Angular velocity for 30 with the horizontal Let Acceleration due to gravity be

Time of travel the ball before impact be t Let the angular velocity for 30 with the horizontal be Vertical distance travelled by the ball is 2ho 27 Let Velocity of the ball before impact with wall be hb Horizontal distance travelled by the ball is ho 2/ h,g hb Let Velocity of the ball after impact with wall be Vha Then from coefficient of restitution formula

ha Substituting h,8 0.85-V2 ha (100 85 V 2 (20 s 17 V 2 ha Velocity of the ball after impact with wall ha is the initial velocity of the ball in the upward travel. At Rebound Height, the final velocity of ball Vf will be zero Conventionally the acceleration is considered negative So using the kinematics equation

400 ( hog 2gh 289J 2 28922g 100h, 289 100/ 289 The rebound height is The angular velocity is given by Herea17 2is the velocity after impact 100h, 289 Substituting

20 h,g 17 289 20 ho8 h,g \( 289 2 100h。 2 5% 17 g The angular velocity for 30 with the horizontal after impact is 17 g