(Solved): Ball A is rolling down on an inclined plane with the velocity of 11 m/s at Point P where the heigh ...

Ball A is rolling down on an inclined plane with the velocity of 11 m/s at Point P where the height from the ground h = 0.8 m as shown in Figure 5. After moving down and reaches the bottom, Ball A made a collision with Ball B, which is in static condition at Point R. The coefficient of restitution between Ball A and Ball B is e = 0.9. The masses of Ball A, Ball B, and the incline angle of the plane are given by mA = 0.5 kg, m² = 0.3 kg, and 0 = 18°. Neglect the friction between balls and the surface of the plane. [Bola A sedang bergolek ke bawah pada satah condong dengan halaju 11 m/s di Titik P di mana ketinggian dari tanah h = 0.8 m seperti ditunjukkan dalam Rajah 5. Selepas bergerak ke bawah dan sampai ke bawah, Bola A membuat perlanggaran dengan Bola B, yang berada dalam keadaan statik di Titik R. Pekali pemulihan antara Bola A dan Bola B ialah e = 0.9. Jisim Bola A, Bola B, dan sudut condong satah diberikan oleh m 0.5 kg, mB = 0.3 kg, dan 0= 18°. Abaikan geseran antara bola dan permukaan satah.] (a) Determine the velocity of Ball A when it reaches the bottom at Point Q. [Tentukan halaju Bola A apabila ia sampai ke dasar di Titik Q.] (10 Marks/ Markah) (b) Find the velocities for Ball A and Ball B after the collision if both balls are moving to the right after the collision. [Cari halaju Bola A dan Bola B selepas perlanggaran jika kedua-dua bola bergerak ke kanan selepas perlanggaran.] (10 Marks/ Markah) Ball A P h 8 m/s ? Q Figure 5 [Rajah 5] Ball B R