Okay, here is my stab at this, I hope it helps!
You know the bullet's initial velocity, V₀ = 450 m/s
You know the final velocity, V = 220 m/s
You also know how long the bullet accelerates (actually decelerates), 14cm, or .14 m
With this information, you learn that you need this equation.
V² = V₀² + 2a (x - x₀), because we have all the information except a, which is the acceleration. So putting it into the equation, it looks like this.
(220m/s)² = (450m/s)² +2a(.14m - 0m)
I'll let you solve the rest, but here are some hints. Your answer will be really big because the bullet slows down really quickly in a really small distance, and you answer will be negative, because this acceleration is causing the bullet to go slower, which is also called deceleration. Hope that helps!
Answer:
Fa=774 N
Fb=346 N
Explanation:
We will solve this problem by equating forces on each axis.
- On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
- On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative
While towing we know that car is mot moving in y-direction so net force in y-axis must be zero
⇒∑Fy=0
⇒
⇒
⇒
Given that resultant force on car is 950N in positive x-direction
⇒∑Fx=950
⇒
⇒
⇒
⇒
⇒
⇒ 
⇒
Therefore approximately, Fa=774 N and Fb=346 N
Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
