Question

A box of mass 3 kg moves on a horizontal frictionless table and collides with another box
of mass 3 kg initially at rest on the edge of the table at height 1 m. The speed of the moving
box just before the collision is 4 m/s. The two boxes stick together and fall from the table.
The kinetic energy just before the boxes strike the floor is (Assume g= 10 m/s2)
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Answer 1

Explanation:

m1 = 3kg

m2 = 3kg

v1 = 4m/s

v2 = m1v1/m1 +m2

= 12/6 = 2m/s

Horizontal initial and final velocity are same

h=1

t =

$t \: = \sqrt{2h \div g}$

= 1/2.2

vertical initail velocity is zero

v = u + gt

= 4.4

total velocity = root v(horizonral)^2 +v(vertical)^2

=

Answer 2

Answer:

hope this is the answer