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igor_vitrenko [27]
3 years ago
9

A 2.0-kg cart is rolling along a frictionless, horizontal track towards a 1.8-kg cart that is held initially at rest. The carts

are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is 5.9 m/s, and the second cart's velocity is -2.7 m/s. (a) What is the total momentum of the system of the two carts at this instant
Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

the total momentum of the system before collision is 6.94 kgm/s

Explanation:

Given;

mass of the first cart, m₁ = 2.0 kg

mass of the second cart, m₂ = 1.8 kg

velocity of the first cart before collision, u₁ = 5.9 m/s

velocity of the second cart before collision, u₂ = -2.7 m/s

The total momentum of the system before collision is calculated as follows;

P_t = P_1 + P_2 \\\\P_t = m_1u_1 + m_2u_2\\\\P_t = (2\times 5.9) + (1.8 \times -2.7)\\\\P_t = 11.8 - 4.86\\\\P_t = 6.94 \ kgm/s

Therefore, the total momentum of the system before collision is 6.94 kgm/s

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List out all the variables that you do know; 

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m 
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as 
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41 
v=√47.41 
v=6.88549 ≈ 6.9 ms⁻¹ 

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Read 2 more answers
Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He
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<h3>Answer:  130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

  • The initial speed (let's call that Vi) is 8.0 m/s
  • The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
  • This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

  • At 0 seconds, he's going 8 m/s
  • At the 1 second mark, he's slowing down to 8-2 = 6 m/s
  • At the 2 second mark, he's now at 6-2 = 4 m/s
  • At the 3 second mark, he's at 4-2 = 2 m/s
  • Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

  • F = force applied
  • m = mass
  • a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

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