Linear Collisions; Conservation Law

AI Thread Summary
The discussion focuses on a collision experiment involving two masses and springs, where mass A (100 g) is propelled by a compressed spring. The student successfully calculated the speed of mass A as it leaves the spring at 0.71 m/s using energy conservation principles. However, while applying momentum conservation to find the velocity of mass B after the collision, the student initially miscalculated the velocity due to a unit conversion error, confusing 20 cm/s with 0.02 m/s. The correct velocity of mass B after the collision is determined to be 0.25 m/s to the right, and the maximum compression of the second spring upon collision is 2.5 cm. The discussion highlights the importance of accurate unit conversions in physics calculations.
trulyfalse
Messages
35
Reaction score
0
Hello Physics Forums!

Homework Statement


The student sets up a collision experiment with two masses on a horizontal air track, with two springs on each end. The mass of A is 100 g and B is 200 g. Each spring has a constant of 20 N/m. Spring 1 (left side) was initially compressed 5 cm
to start mass A moving.

a. Determine the speed of mass A as it leaves spring 1 [0.71 m/s]
b. If A has a velocity of 20 cm/s to the right after it collides with B, determine the velocity of mass B. [0.25 m/s right]
c. Determine the maximum compression of spring 2 when mass B collides with it. [2.5cm]

Homework Equations


Ep = 1/2kx2
p = mv

The Attempt at a Solution


I solved part a readily using the conservation of energy law (Ep = Ek). When I tried to solve part b using conservation of momentum, however, I obtained an incorrect answer:

Total momentum before = Total momentum after
m1v1+m2v2=m1v3+m2v4
(m1v1+m2v2-m1v3)/m2=v4
[(0.1kg)(0.71m/s)+(0.2kg)(0)-(0.1kg)(0.02m/s)]/(0.2kg)= v4
v4 = 0.35 m/s right

Am I reading the question wrong? Are my formulae incorrect?​
 
Last edited:
Physics news on Phys.org
20 cm/s would be 0.20 m/s, not 0.02 m/s
 
Hahahaha thanks for your help; I'll have to be more diligent with unit conversions next time!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

Sequental elastic and inelastic collisions
Replies
4
Views
2K
High School Physics Question -- Skateboarder Jumps & Conservation of Momentum
Replies
3
Views
3K
How do I solve for an elastic collision going up a ramp?
Replies
10
Views
3K
Elastic collision of block on block
Replies
4
Views
2K
Ambiguous question on momentum and how it works...
Replies
13
Views
1K
How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?
7
Replies
335
Views
14K
Finding the Line of Motion of Two Particles
Replies
16
Views
1K
SHM and Conservation of Energy/Momentum
Replies
1
Views
2K
Problem with when to use Force and Energy. What compresses spring/
Replies
3
Views
1K
Find elastic energy in the compressed spring.
Replies
12
Views
3K

Hot Threads

Recent Insights

Back
Top