The book that weighs less
Answer:
-92.33 (meaning the objects will not meet above the ground).
Explanation:
We can use the kinematic equation <em>displacement = initial velocity*time + 1/2*acceleration*time^2.</em>
We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:
x = 0*t + 1/2*(-9.8)*t^2+45
x = 8.5*t + 1/2*(-9.8)*t^2
We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:
0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2
-4.9*t^2 +45 = 8.5*t + -4.9*t^2
45 = 8.5*t
t = 45/8.5 ≈5.294
Now, we can plug t as 5.294 into any of the equations above to solve for x:
x = 0*5.294 + 1/2*-9.8*(5.294)^2+45 ≈ -92.33
That means, the objects will not meet above the ground.
Explanation:
Given:
Final speed of mass A = Va
Final speed of mass B = Vb
Mass of A = Ma
Mass of B = Mb
Ma = 2 × Mb
By conservation of linear momentum,
0 = Ma × Va + Mb × Vb
0 = 2 × Mb × Va + Mb × Vb
Vb = - 2 × Va
Energy of the spring, U = 1/2 × k × x^2
1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
Ma × Va² + Mb × Vb² = 70
2 × Mb (-Vb/2)² + Mb × Vb² = 70
1/2 × Mb × Vb² + Mb × Vb² = 70
3/2 × Mb × Vb² = 70
Mb × Vb² = 140/3
= 46.7 J
Ma = 2 × Mb and Vb = - 2 × Va
Ma/2 × (4 × Va²) = 140/3
Ma × Va² = 70/3
Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J
Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J
We know, F = k * q₁ * q₂ / r²
Substitute the known values,
F = 9 * 10⁹ * 5 * 7 / (1.2)²
F = 315 * 10⁹ / 1.44
F = 218.75 * 10⁹ N
F = 2.1875 * 10¹¹ N [ Final Answer ]
Hope this helps!
