Mechanics Momentum/spring energy question

[Mathy]

Hey as i have had some help with my thermodynamics why not my mechanics too. I have a question of four parts, a b c d, i have worked out a b and d but can't do c

A 2kg sphere attached to a light arm or length 0.5m is released from a horizontal postition to hit a 10kg carriage which is backed up by the spring with stiffness of 1kn/m. the carriage is initally at rest with the spring uncompressed. assuming elastic collision calculate

a: the speed of the sphere (3.13m/s worked out)
b: the rebound speed (-2.09m/s becasue of direction, worked out)
c:the maximum tension in the light arm after impact? i don't know
d: the maximum travel of the carriage after impact, (0.1m)

we also know speed of the carriage to be 1.04m/s

sorry i can't get the diagram up, don't know how to put it in code

carriage is on right with the sroing behind it, the pendulum swinging down to the right and rebounding left.

 

[Mathy]

oesnt matter i have done it i was doing t-mg=ma but in this case its t+mg=ma, so no need to reply

 

[piercas]

Hello,

Thank you for reaching out for help with your mechanics question. I can see that you have already made good progress in solving parts A, B, and D. I will now provide a solution for part C.

To calculate the maximum tension in the light arm after impact, we need to consider the conservation of momentum and energy. The initial momentum of the 2kg sphere is zero since it is released from a horizontal position. After the collision, the sphere and the carriage will move together with a combined mass of 12kg. This means that the final velocity of the combined system will be:

vf = (m1v1 + m2v2) / (m1 + m2) = (2kg)(3.13m/s) + (10kg)(1.04m/s) / (12kg) = 1.65m/s

Now, using the conservation of energy, we can calculate the maximum tension in the light arm. The energy before the collision is all in the form of gravitational potential energy, which is given by mgh. After the collision, the energy is in the form of kinetic energy, given by 1/2mv^2. Since the sphere and the carriage are moving together, we can equate these two energies to find the maximum tension:

mgh = 1/2mv^2

Solving for h, we get:

h = v^2 / (2g)

Plugging in the values for v and g, we get:

h = (1.65m/s)^2 / (2)(9.8m/s^2) = 0.14m

This is the maximum height that the sphere will reach after the collision. The maximum tension in the light arm will occur at this point, and it can be calculated using the equation for the tension in a spring:

T = kx

Where k is the stiffness of the spring and x is the maximum displacement of the spring. In this case, the maximum displacement is equal to the maximum height, so we have:

T = (1kn/m)(0.14m) = 0.14kN

Therefore, the maximum tension in the light arm after impact is 0.14kN. I hope this helps you in solving your mechanics question. Keep up the good work!