Question

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder had been placed on one of the blocks. The explosion does not harm the blocks but it doubles their total kinetic energy. After the explosion the blocks move along the x axis and the incident block?

Answer 1

Speed of 1.83 m/s and 6.83 m/s

Step-by-step explanation:

From the principle of conservation of momentum

`mv_o=m(v_1 + v_2)` where m is the mass,
`v_o` is the initial speed before impact,
`v_1` and
`v_2` are velocity of the impacting object after collision and velocity after impact of the originally constant object

`5m=m(v_1 +v_2)`

Therefore
`v_1+v_2=5`

After collision, kinetic energy doubles hence

`2m*(0.5mv_o)=0.5m(v_1^(2)+v_2^(2))`

`2v_o^(2)=v_1^(2) + v_2^(2)`

Substituting 5 m/s for
`v_o` then

`2*(5^(2))= v_1^(2) + v_2^(2)`

`50= v_1^(2) + v_2^(2)`

Also, it’s known that
`v_1+v_2=5` hence
`v_1=5-v_2`

`50=(5-v_2)^(2)+ v_2^(2)`

`50=25+v_2^(2)-10v_2+v_2^(2)`

`2v_2^(2)-10v_2-25=0`

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then
`v_2=6.83 m/s`

Substituting,
`v_1=-1.83 m/s`

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s