What is the Archer's Velocity After Firing Two Arrows on Frictionless Ice?

In summary: Just check your units and then you should be fine. In summary, the 68.6 kg archer will have a velocity of 54.8 m/s after firing the second arrow.
  • #1
aaronb
23
0

Homework Statement


A 68.6 kg archer stands at rest on frictionless ice and fires a 0.46 kg arrow horizontally. He fires and slides back -0.37m/s. He fires another arrow while he is still sliding that is going 76m/s. What will be his velocity after he fires the second arrow?

Homework Equations


m1v1f + m2v2f = -0.37

The Attempt at a Solution


I solved for v1f and plugged in the numbers to get

v1f = -.37/68.6 - .46(76)/68.6 = -.515
 
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  • #2
aaronb said:

Homework Statement


A 68.6 kg archer stands at rest on frictionless ice and fires a 0.46 kg arrow horizontally. He fires and slides back -0.37m/s. He fires another arrow while he is still sliding that is going 76m/s. What will be his velocity after he fires the second arrow?

Homework Equations


m1v1f + m2v2f = -0.37

The Attempt at a Solution


I solved for v1f and plugged in the numbers to get

v1f = -.37/68.6 - .46(76)/68.6 = -.515

That doesn't look quite right. Just check your units. You have an equation in which momentum = velocity?

Maybe figure the speed at which the initial arrow was shot? And then the total momentum of both arrows will yield the reaction momentum of the archer and hence his velocity?
 
  • #3
since the archer is moving -0.37m/s should I multiply that by the mass of the archer so I get momentum on the right side of the equation? The velocity of the first arrow is 54.8 m/s
 
  • #4
aaronb said:
since the archer is moving -0.37m/s should I multiply that by the mass of the archer so I get momentum on the right side of the equation? The velocity of the first arrow is 54.8 m/s

I think it amounts to the same thing. Whether you start with no momentum and then figure the sum of the 2 momenta as one, or figure them serially.

There is the more subtle treatment of the mass of the arrows as to whether he was holding the second when he shot the first and hence his mass is reduced with the second shot. But I wouldn't put that fine a point on it.
 

1. What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is defined as the product of an object's mass and velocity. In other words, it is a measure of how difficult it is to stop an object from moving.

2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass, measured in kilograms (kg), by its velocity, measured in meters per second (m/s). The equation for momentum is p = m * v, where p is momentum, m is mass, and v is velocity.

3. What is an isolated system?

An isolated system is a physical system that does not exchange any matter or energy with its surroundings. This means that the total momentum of an isolated system remains constant, as there are no external forces acting on the system to change its momentum.

4. How does momentum relate to Newton's Laws of Motion?

Momentum is directly related to Newton's Laws of Motion. According to Newton's First Law, an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. This means that an object's momentum will remain constant unless acted upon by an external force. Additionally, Newton's Second Law states that the net force on an object is equal to its mass multiplied by its acceleration, which is essentially the change in momentum over time.

5. Can momentum be transferred between objects?

Yes, momentum can be transferred between objects. This is known as an elastic or inelastic collision. In an elastic collision, the total momentum of the system is conserved, meaning that the total momentum before and after the collision remains the same. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound, but the total momentum of the system is still conserved.

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