# What is the maximum distance the frame moves when a putty is dropped onto it?

• Swatch
In summary, the frame stretches the spring 0.050 m and the putty is dropped from a height of 30.0 cm. The frame moves downward an maximum of 23.2 cm.
Swatch
A 0.150 kg frame when suspended from a spring, stretches the spring 0.050 m. A 0.200 kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm. I have to find the maximum distance the frame moves downward from its initial position.

I have calculated the spring constant to be 1.47N/0.050m = 29.4 N/m

The first stage is when the putty is falling down or:
Vp2 = sqrt(0.6*g)

Then I'm not sure. Can I say that momentum is conserved during the collision of the putty and frame?

The last stage would be conservation of energy when putty and frame travel downwards. But I need to know the total speed.

Could someone please give me a hint

Yes use conservation of momentum to find the speed of the system putty+frame.

But since there is a force of gravitation can I really use conservation of momentum?

Yes, it would be an impulse approximation. During the collision the force will be greater than the magnitude of the gravitation force, because the time of collision will be small.

O.K. What I've done is this.

For the spring I calculated the spring constant

K=1.47N/0.050m = 29.4 N/m

For the falling putty I found the final speed of the putty just before it hits the frame.
Vp2 = sqrt(g*0.6)

The collision of the putty and frame is
Vt = speed of putty and frame

Vt = mpVp2/mt
(where mp is the mass of the putty, mt is the mass of the putty and frame).

Then when the frame and putty are traveling downwards I use conservation of energy

Ug2 + Ue2 = K1 + Ug1 + Ue1 - K2
(Ug =gravitational potential energy, Ue = elastic potential energy)

Here I end up with:

y(squared)*k*0.5 + y*g*mt = 0.5mp(squared) * g*0.6/mt(squared) + 0.0367

From this I get y = (-3.43 +- sqrt(11.8+10.7) )/29.4

I should get 23.2 cm But I always get a wrong answer.

Does anyone see what I'm doing wrong?

I can't quite follow the details of your calculations, but when you apply conservation of energy after the collision be sure to take into account that the spring has an initial elastic potential energy since it is stretched. You are trying to find the amount of additional stretch.

After the collision I did:

Ug2 + Ue2 = K1 + Ug1 + Ue1 - K2
(Ug =gravitational potential energy, Ue = elastic potential energy)

mt*g*y + 0.5*k*y(squared) = 0.5*mt*vt(squared) + 0 + 0.5*k*0.050(squared) - 0

where k is the spring constant

I take the origin to be where the spring in unstreched and the stretched position as -0.050m.

What about Ug1? If the origin is the unstretched postion, then Ug1 is not zero.

Ug1 was the problem. Thanks Doc Al.

## 1. What is momentum and how is it calculated?

Momentum is a measure of an object's motion and is calculated by multiplying its mass and velocity. The formula for momentum is p = m * v, where p is momentum, m is mass, and v is velocity.

## 2. What is the difference between linear momentum and angular momentum?

Linear momentum refers to the motion of an object in a straight line, while angular momentum refers to the motion of an object around a fixed axis. Linear momentum is calculated by multiplying mass and velocity, while angular momentum is calculated by multiplying moment of inertia and angular velocity.

## 3. How does momentum relate to Newton's laws of motion?

Newton's first law states that an object in motion will remain in motion unless acted upon by an external force. This is related to momentum because an object with momentum will continue to move unless a force is applied to change its velocity. Newton's second law states that force is proportional to mass and acceleration, which can be rewritten as force = mass * change in velocity / time. This is essentially the same formula as momentum, showing the strong relationship between the two.

## 4. Can momentum be conserved in a closed system?

Yes, momentum is conserved in a closed system where there are no external forces acting on the objects. This is known as the law of conservation of momentum, where the total momentum before an event is equal to the total momentum after the event.

## 5. How is momentum used in real-world applications?

Momentum is used in various real-world applications, such as in car safety features like airbags, where the force of the impact is reduced by increasing the time taken for the car to stop. It is also used in sports, such as in football, where players use momentum to gain an advantage over their opponents. Momentum is also important in space travel, as rockets use their momentum to propel themselves forward.

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