Answer:
Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J
max spring compression where both velocities are the same: conserve momentum:
1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s
which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J
The remaining energy went into the spring:
U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x²
x = 0.0076 m ↠(a)</span>
Answer:
The ratio is KE : TM = 0.75
Explanation:
from the question we are told that
The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position
Generally the total mechanical energy of the mass is mathematically represented as

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring
Generally this total mechanical energy is mathematically represented as

=> 
Here the potential energy of the mass is mathematically represented as
![PE = \frac{1}{ 2} * k * [ x ]^2](https://tex.z-dn.net/?f=PE%20%20%20%3D%20%5Cfrac%7B1%7D%7B%202%7D%20%20%2A%20%20k%20%2A%20%20%5B%20x%20%5D%5E2)
Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

So
![PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2](https://tex.z-dn.net/?f=PE%20%20%20%3D%20%5Cfrac%7B1%7D%7B%202%7D%20%20%2A%20%20k%20%2A%20%20%5B%20%5Cfrac%7BA%7D%7B2%7D%20%20%5D%5E2)
So
![KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2](https://tex.z-dn.net/?f=KE%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20A%5E2%20-%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20%5B%5Cfrac%7BA%7D%7B2%7D%20%5D%5E2)
=> 
=> 
So the ratio of
is mathematically represented as

=>
Answer:
0.4778 m/s
Explanation:
To solve this question, we will make use of law of conservation of momentum.
We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;
V_x = (12 m/s)(cos(35°)) = 9.83 m/s.
Thus, the horizontal component of the rock's momentum is;
(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.
Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.
Thus, to get the person's speed, we know that; momentum = mass x velocity
Mass of person = 72 kg and we have momentum as 34.405 kg·m/s
Thus;
34.405 = 72 x velocity
Velocity = 34.405/72
Velocity = 0.4778 m/s
Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan