Question

Consider Figure P8.82 in the book on p. 268.

a) Solve Problem 8.82 as stated;
b) solve the same problem but for the case of a completely elastic collision;
​​​​​​​c) solve the same elastic case as in (b) but for the situation that the mass initially at rest at the bottom of the bowl is twice that of the mass that is released

8.82 - CP Two identical masses are released from rest in a smooth hemispherical bowl of radius R from the positions shown in Fig. P8.829. 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? Figure P8.82

Answer 1

speed of 1 just before collision, u, ſagr Now, if after collision they move with speed to e Then conservation line ar momentum gives as mu, amo -> vnu, Suppose they reach to height hi Then we write using energy conservation principle het zwagh = ha estat (1987). 29R can 6 In conoce In case of elastic collision of two identical particles, their velocities are en changed. so, I will stop after collision and a will nome nove with speed " . Suppose 2 reaches to height h. Then we can write using conservation of energy: mgha muza mura / m. Rgr a [har In case of elastic colle sien of two bodies, velocités after collision are given as 2, m,-me ) e + 2mqų mitma mitm. ve- (m.-m, ) U + am, u, . + mit me putting - u= /292 y =o & m,am & m,:am we gut U2= (2m-m) xo 2m W29R. ²2 gr almth mtam so, height reached will be . home una by a 4x78R 9x28 12