# Linear Collisions; Conservation Law

• trulyfalse
In summary, a student sets up a collision experiment with two masses on a horizontal air track. Mass A is 100 g and B is 200 g, with each spring having a constant of 20 N/m. Spring 1 is compressed 5 cm to the left to start mass A moving. Part a asks for the speed of mass A as it leaves spring 1, which is 0.71 m/s. Part b asks for the velocity of mass B after colliding with A, which is 0.25 m/s to the right. Part c asks for the maximum compression of spring 2 when mass B collides with it, which is 2.5 cm. The student made a mistake in their attempt at

#### trulyfalse

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]

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:
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!