<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.
<h3>What is the relation between the masses of A and B?</h3>
Mass of piece B = Mb
- Velocities of pieces A and B are Va and Vb respectively.
- As per conservation of momentum,
Ma×Va = Mb×Vb
So, 1.9Mb × Va = Mb×Vb
=> 1.9Va = Vb
<h3>What are the kinetic energy of piece A and B?</h3>
- Expression of kinetic energy of piece A = 1/2 × Ma × Va²
- Kinetic energy of piece B = 1/2 × Mb × Vb²
- Total kinetic energy= 7900J
=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900
=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900
=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j
=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule
- Kinetic energy of piece B = 7900 - 2724 = 5176 Joule
Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.
Learn more about the kinetic energy here:
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Answer:
The speed of the 11.5kg block after the collision is V≅4.1 m/s
Explanation:
ma= 4.8 kg
va1= 7.3 m/s
va2= - 2.5 m/s
mb= 11.5 kg
vb1= 0 m/s
vb2= ?
vb2= ( ma*va1 - ma*va2) / mb
vb2= 4.09 m/s ≅ 4.1 m/s
Answer:
its speed when its height was half that of its starting point is 25.46 m/s
Explanation:
Given;
final speed of the roller coaster, v = 36 m/s
Applying general equation of motion;
V² = U² + 2gh
where;
V is the final speed of the roller coaster
U is the initial speed of the roller coaster = 0
h is the height attained at a given velocity
36² = 0 + (2 x 9.8)h
1296 = 19.6 h
h = 1296/19.6
h = 66.1224 m
when its height was half that of its starting point, h₂ = ¹/₂ h
h₂ = ¹/₂(66.1224 m) = 33.061 m
At h = 33.061 m, V = ?
V² = U² + 2gh
V² = 0 + 2 x 9.8 x 33.061
V² = 648
V = √648
V = 25.46 m/s
Therefore, its speed when its height was half that of its starting point is 25.46 m/s