Question

6. A block of mass 100 kg slides down a frictionless surface of height 30 cm and collides elastically with another block of mass 5 kg, and falls of a table. If the table has a height of 90 cm, where does each block land?

Answer 1

velocity of 100 kg just before impact

u^2 = 2 g h = 2 * 9.8* 0.3

u = 2.425 m/s

now velocity of 100 kg after strike

v(100) = (m1 - m2) u 1 / ( m1 + m2) = ( 100 - 5) * 2.425 / ( 100 + 5)

v(100) = 2.19 m/s

velocity of 5 kg after strike

v(5) = 2 m1 u1 / ( m1 + m2) = 2* 100* 2.425 / ( 5 + 100) = 4.619 m/s

now time taken by each to reach ground

h' = 0.5 gt^2

0.9 = 4.9 t^2

t = 0.4286 s

now,

horizontal distance travelled by 100 kg

D(100) = v(100) * t = 2.19* 0.4286 = 0.939 m

horizontal distance traveled by 5 kg

D(5) = 4.619 * 0.4286 = 1.98 m

=====

Comment before rate in case any doubt, will reply for sure.. goodluck