I think the answer is c. the more characteristics they have in common. Sorry if wrong.
Answer:
x_total = 20m
Explanation:
This is an exercise in kinematics, we will look for the distance it travels during the reaction time, where there is no braking, and then the distance during the deceleration.
Distance traveled during response time
v = x₁ / t
x₁ = v t
let's calculate
x₁ = 10 0.75
x₁ = 7.5 m
Now we calculate the distance during braking, the final speed is zero (v = 0)
v² = v₀² - 2 a x₂
x₂ = v₀² / 2 a
let's calculate
x₂ = 10² / (2 4)
x₂ = 12.5m
the total stopping distance is
x_total = x₁ + x₂
X_total = 7.5 + 12.5
x_total = 20m
What are the answers
Matter is made of mass
Answer:
![v_2=24\ m/s](https://tex.z-dn.net/?f=v_2%3D24%5C%20m%2Fs)
Explanation:
It is given that,
Mass of puck 1, ![m_1=0.5\ kg](https://tex.z-dn.net/?f=m_1%3D0.5%5C%20kg)
Mass of puck 2, ![m_2=2\ kg](https://tex.z-dn.net/?f=m_2%3D2%5C%20kg)
Initial speed of puck 1, ![u_1=80\ m/s](https://tex.z-dn.net/?f=u_1%3D80%5C%20m%2Fs)
Initial speed of puck 2, ![u_2=0\ m/s](https://tex.z-dn.net/?f=u_2%3D0%5C%20m%2Fs)
After the collision, the speed of puck 1, ![v_1=-16\ m/s](https://tex.z-dn.net/?f=v_1%3D-16%5C%20m%2Fs)
Let
is the final velocity (in m/s) of puck 2 after the collision. Using the conservation of momentum as :
![m_1u_1+m_2u_2=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2)
![0.5\times 80+2\times 0=0.5\times (-16)+2v_2](https://tex.z-dn.net/?f=0.5%5Ctimes%2080%2B2%5Ctimes%200%3D0.5%5Ctimes%20%28-16%29%2B2v_2)
![v_2=24\ m/s](https://tex.z-dn.net/?f=v_2%3D24%5C%20m%2Fs)
So, the final velocity of the puck 2 after the collision is 24 m/s. Hence, this is the required solution.
Work done = change in PE = mgh = 60*10*4 = 2400 J