Linear Momentum and Collisions- Acceleration Relative to the Ice

[rsfancy]

On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.10 kg sled is pulled with a horizontal force of 37.0 N, it begins to move with an acceleration of 2.50 m/s^2.The 23.0 kg child accelerates too, but with a smaller acceleration than that of the sled. Thus, the child moves forward relative to the ice, but slides backward relative to the sled.

Q.Find the acceleration of the child relative to the ice.

There are several equations that fall under the Linear Momentum and Collision category and none of them seem to quite be relevant enough to the question, obviously this is mostly due to the fact that I don't see the correlation. None of the formulas seem to help me to find an Acceleration, neither does Energy Conservation( which is what I was thinking would help in this problem) help me to find Acceleration.

I am a little confused as to how to go about this problem as I am unable to see which way to even start, or what I should be using to get to the end result, the Acceleration of the child relative to the ice.

Any help or guidance would be much appreciated.

 

[Stovebolt]

On a cold winter morning, a child sits on a sled resting on smooth ice. When the 9.10 kg sled is pulled with a horizontal force of 37.0 N, it begins to move with an acceleration of 2.50 m/s^2.The 23.0 kg child accelerates too, but with a smaller acceleration than that of the sled. Thus, the child moves forward relative to the ice, but slides backward relative to the sled.

Q.Find the acceleration of the child relative to the ice.

There are several equations that fall under the Linear Momentum and Collision category and none of them seem to quite be relevant enough to the question, obviously this is mostly due to the fact that I don't see the correlation. None of the formulas seem to help me to find an Acceleration, neither does Energy Conservation( which is what I was thinking would help in this problem) help me to find Acceleration.

I am a little confused as to how to go about this problem as I am unable to see which way to even start, or what I should be using to get to the end result, the Acceleration of the child relative to the ice.

Any help or guidance would be much appreciated.

The best way to start almost any physics problem is to draw a picture of the problem and then to make a Free Body Diagram of each component. Think about each force that might be present and think about what the sum of all forces will be. See if your diagrams give you any ideas.

 

[baba89]

I would approach this problem by first understanding the concept of relative motion. In this case, the child is experiencing two different accelerations, one relative to the sled and one relative to the ice. To find the acceleration of the child relative to the ice, we need to use the concept of the relative velocity between the child and the sled.

We can start by defining some variables:
- m1 = mass of the sled (9.10 kg)
- m2 = mass of the child (23.0 kg)
- F = horizontal force applied to the sled (37.0 N)
- a1 = acceleration of the sled (2.50 m/s^2)
- a2 = acceleration of the child relative to the sled (unknown)
- a2' = acceleration of the child relative to the ice (unknown)

We can use Newton's second law of motion to relate the forces and accelerations:
F = m1a1
37.0 N = (9.10 kg)(2.50 m/s^2)
a1 = 4.07 m/s^2

Next, we can use the concept of relative motion to relate the accelerations of the child and the sled:
a2 = a2' + a1
a2' = a2 - a1
We know that the sled is accelerating at 2.50 m/s^2 and the child is accelerating at a smaller rate, so we can say that a2 < a1. This means that the child is moving forward relative to the ice, but at a slower rate than the sled.

Now, we can use the concept of inertia to relate the accelerations and masses of the sled and the child:
m1a1 = m2a2
(9.10 kg)(2.50 m/s^2) = (23.0 kg)a2
a2 = 0.924 m/s^2

Finally, we can substitute this value into our equation for relative motion to find the acceleration of the child relative to the ice:
a2' = a2 - a1
a2' = 0.924 m/s^2 - 4.07 m/s^2
a2' = -3.15 m/s^2

This negative acceleration means that the child is sliding backwards relative to the sled, but moving forward relative to the ice at a slower rate than the sled. Therefore, the acceleration