Question

A 50 kg person stands on a 25 kg sled. The sled and person are initially moving together at 6.0 m/s in the +x direction on a frictionless, horizontal surface. If the person jumps off the back of the sled and now slides at 2.0 m/s in the +x direction, how fast is the sled moving?

Answer 1

The sled is moving at 14 m/s

Step-by-step explanation:

Using the law of conservation of momentum, we know that:

Total Momentum of system before collision = Total Momentum of system after collision

(m1)(u1) + (m2)(u2) = (m1)(v1) + (m2)(v2)

Here,

m1 = mass of the person = 50 kg

m2 = mass of sled = 25 kg

u1 = initial speed of the person = 6 m/s

u2 = initial speed of sled = 6 m/s

v1 = final speed of the person = 2 m/s

v2 = final speed of sled = ?

Therefore,

(50 kg)(6 m/s) + (25 kg)(6 m/s) = (50 kg)(2 m/s) + (25 kg)(v2)

(450 kg.m/s - 100 kg.m/s)/25 kg = v2

v2 = 14 m/s

Answer 2

The sledge final speed is of 8 m/s.

Step-by-step explanation:

Considering the sledge-person system and due there are no forces applying over the system, then the momentum conservation principle can be used:

[tex] F=dP/dt=Pf-Pi=0[/text]

Where dP/dt is the momentum variation in time between the final f and initial i states respectively. As the variation is zero, then the momentum is constant:

[tex] Pf=Pi[/text] (1)

From the momentum definition we know:

[tex] P=m.v[/text]

Where the product is a scalar product since the velocity v is a vector. The variable m represents the mass.

For equation (1), Pi and Pf are defined as:

[tex] Pi =ms.vs_{i}+mp.vp_{i}[/text] (2A)

[tex] Pf =ms.vs_{f}+mp.vp_{f}[/text] (2B)

Where the subcrpit p and b relate to the sled and the person respectively. Note that for the initial momentum, the velocity are the same for the person and the sledge:

[tex] Pi =(ms+mp)v_{i}[/text] (3A)

On the final momentum equation (2B), the final sled velocity it is known but the person velocity is unknown. By replacing equation (3A), and (2B) in equation (1):

[tex] ms.vs_{f}+mp.vp_{f} =(ms+mp)v_{i}[/text]

By solving for the final sledge velocity:

[tex] vs_{f} =((ms+mp)v_{i}-(ms.vs_{f}))/mp[/text]

Finally, replacing the respective values:

[tex] vp_{f} =((50 kg +25 kg)*6 m/s-(25 kg * 2 m/s))/(25 kg)[/text]

[tex] vp_{f} =((450 kg.m/s-(50 kg .m/s))/(50 kg)[/text]

[tex] vp_{f} =(400 kg .m/s)/(50 kg) = 8 m/s[/text]