Since
<span>Vf^2 = 2*a*S </span>
<span>Given S=3.6m, thus </span>
<span>a = Vf^2/(2*3.6) </span>
<span>a = Vf^2/7.2 </span>
<span>Let d be the distance along the slope at which the velocity is 0.5Vf, then </span>
<span>(0.5Vf)^2 = 2*a*d </span>
<span>or </span>
<span>d = (0.5*Vf)^2/(2*a) </span>
<span>with a = Vf^2/7.2, we have </span>
<span>d = 0.9 m</span>
A pendulum is probably the most common showing of this example. As the pendulum swings down, it converts its potential energy (height) into kinetic energy (velocity). At the lowest point the kinetic energy is the highest and the potential is the lowest. At the highest point in its swing the velocity is zero so the kinetic energy is zero and the potential energy is at a maximum (greatest height).
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
Answer:
- Here we use the conservation of momentum theorem.
- m stands for mass, and v stands for velocity. The numbers refer to the respective objects.
- m1v1 + m2v2 = m1vf1 + m2vf2
- Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.
- m1v1 + m2v2 = vf(m1 + m2)
<u>Let’s substitute in our givens.</u>
(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)
I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.
Note that I have considered the bullet’s velocity to be in the positive direction,
The answer is vf = 0.280 m/s
Yes if you need to know why i will tell you why
