CP? Two identical masses are released from rest in a smooth hemispherical bowl of radius R from the positions shown in ?Fig. P8.82.? Ignore friction between the masses and the surface of the bowl. If the masses stick together when they collide, how high above the bottom of the bowl will they go after colliding?
Solution 86 P Step 1: Let us consider the figure given From this we can notice that one block is at the top of left side of arc while the other is at the center We shall find the energy required by the one block to collide with other Energy of block one E1= mgR-----------(1) Velocity of block of B is B 2 And v B 1/2 mv B ---------(2) Using (1) and (2) We get v = 2gR B After the collision linear momentum is conserved Hence the velocity of the system after collision will be v = 2gR = mv B 2mv A Hence we get velocity of particle A as vA= v B2 Substitute forB= 2gR We get vA= gR/2 Velocity of particle after collision will be 2 = 1/2 2mv A This can further written as = 2mg y (y = change in position along y axis ) To find the position of the block after collision y = v A2g But we have vA= gR/2 Hence we can write as y = gR/2g × 2 Simplifying this get y = R/4 Hence we get the position of the block after the collision as R/4