Solving Collision & Rotation Q: Find Impulse, Force, Speed, Length

[Lord Dark]

Homework Statement

hi guys,, got a question about collision + rotation ,, the following question is :
A steel ball of mass 0.5 kg is attached to a cord that is fixed at the other end. The ball
is then released when the cord is horizontal, as shown in the figure. At the bottom of its
path, the ball elastically strikes a 2 kg steel block initially at rest on a frictionless
horizontal surface. The ball is in contact with the block for 2 ms, and the force due to the
elastic collision is given by:
F(t) = 8×106t – 2×109t^2
for 0 ≤ t ≤ 4 ms, where t is the time in seconds. Both the ball and block are considered as
point particles.

Find:
a) the impulse on the block due to the collision.
b) the average force of the block from the ball during the period of contact.
c) the maximum force on the block during the period of contact.
d) the speed of block immediately after the collision and the length of the cord.

Homework Equations

The Attempt at a Solution

i know that J=Favg delta(T) but i don't know either of them and i don't know how to answer the first question ,, i need a way to answer it please so i can continue solving the others

 

[Dweirdo]

I'f I'm not wrong You should integrate the force equation, aka calculate the area under the Graph in bounds 2,0 also convert to seconds.

 

[nlightner]

To solve this problem, we can use the principles of conservation of momentum and conservation of energy. We will also need to use the equations for impulse, force, and velocity.

a) To find the impulse on the block due to the collision, we can use the equation J = mΔv, where J is the impulse, m is the mass of the object, and Δv is the change in velocity. In this case, we can calculate the change in velocity of the block by using the conservation of momentum equation, m1v1 = m2v2, where m1 and v1 are the mass and velocity of the ball, and m2 and v2 are the mass and velocity of the block after the collision. We can rearrange this equation to solve for v2, which will give us the change in velocity of the block. Once we have this, we can plug it into the equation for impulse to find the value.

b) To find the average force of the block from the ball during the period of contact, we can use the equation Favg = Δp/Δt, where Δp is the change in momentum and Δt is the time of contact. We can find the change in momentum by using the same equation as in part a, and we can use the given time of contact to find the average force.

c) To find the maximum force on the block during the period of contact, we can use the given equation for force, F(t) = 8×106t – 2×109t^2, and plug in the time of contact to find the maximum value.

d) To find the speed of the block immediately after the collision, we can use the equation m1v1 = m2v2, as mentioned in part a. We can also use the conservation of energy equation, Ei = Ef, where Ei is the initial energy (kinetic and potential) and Ef is the final energy. We can set Ei equal to the initial kinetic energy of the ball and Ef equal to the final kinetic energy of the block, which we can solve for using the equation for velocity. Once we have the final velocity, we can use it to find the length of the cord by using the equation for circular motion, v^2 = ω^2r, where ω is the angular velocity and r is the radius of the circle. We can solve for r to