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 boxes stick together and fall from the table. The kinetic energy just before the boxes strike the floor is (Assume, acceleration due to gravoty, `g=10 m//s^(2)`)

Options

A. 40 J
B. 80 J
C. 96 J
D. 72 J

Answer

D

Explanation

The given situation can be be shown as below,

According to law of conservation of momentum, total momentum before collision = total
momentum after collision.
`m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))v`
Given, mass of boxes, `,_(1)=m_(2)=3 kg`
speed of the moving box, `u_(1)=4` m/s
and initial speed of second box, `u_(2)=0`
Putting the given values, we get
`3xx4+3xx0=(3+3)v`
`implies v=12/6=2` m/s
Thus, the two bodies mover with the velocity of 2 m/s. Applying law of conservation of energy at height of 1 m and at the bottom of table.
`KE_(1)+PE_(1)=KE_(2)+PE_(2)`
At the bottom just above the ground, the total energy is KE as height is approximately equals to zero.
`1/2 (m_(1)+m_(2))v^(2)+(m_(1)+m_(2))g h=KE_(2)`
`implies KE_(2)=1/2 xx6xx4+6xx10xx1`
`=12+60=72 J`
Hence, the kinetic energy just before the boxes strike the floor is 72 J.