Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir
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Answer:
Speed of 1.83 m/s and 6.83 m/s
Explanation:
From the principle of conservation of momentum
where m is the mass,
is the initial speed before impact,
and
are velocity of the impacting object after collision and velocity after impact of the originally constant object
Therefore
After collision, kinetic energy doubles hence
Substituting 5 m/s for then
Also, it’s known that hence
Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then
Substituting,
Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s
We are given: Final velocity ()=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion
Let us plug values of final velocity, and time in above equation.
Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion
Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,
Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.