Question

Result rounded to two decimal points

In a physics laboratory a small cube (M) slides down a frictionless plane as shown in the figure, and elastically strikes another cube (m) that is at the bottom and has only half the mass of the First. If the plane is 35 cm high and the table is 95 cm high measured from the floor [Hint both come out of the plane (table moving horizontally] 35 cm 95 cm a) Where (in m) will the cube of mass M (XM) fall? b) Where (in m) will the cube of plus m (X.) fall?

Answer 1

Given:

• M = The mass of cube 1
• m = The mass of cube 2 = M/2
• h = The height of the incline = 35 cm =0.35 m
• H = The height of table =95 cm = 0.95 m

3 As, m = IM 2 2 I MVA 2 Mgn = *= n2gh = x 2 x 9.8 x 0.35 VA= 2.619 mis Using law of Conservation, 3 M VA = M VA & m vel VA' + m (vazvel] 2 VA = » VAT VA M >> 2VA & VA+VA VA 2.619 3 3 VA'= 0.873 m/s 0.873 m/s & Now, As VB VA+VA' VB 2.619 to.073 3.492 m/s VA

Э We have, Я. 2H ДІЯ Лха- VA + — Д 29, 2 и { 3 А - Ihr 5 ЛХА - 2 o, 35 Храб О. 2594 м). 3 imilarly Ако е в О.35х0, 45 2 3 BnB = 1.53m - (а) Cube fall 1 — то м (Хм) О. 384 m). S (b) Cub 8 mas m (Xm) face all :- 1, 531 м