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?

The sledge final speed is of 8 m/s.

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]

The sled is moving at 14 m/s

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