Elastic collision in one dimension

[lone21]

Homework Statement

A Pendulum weighs .5kg and has a string length of 70cm swings from a horizontal position downwards to hit a block that weighs 2.5 kg and is on a frictionless plain. Calculate the speed of both the ball and the block after the elastic collision.

Homework Equations

mgh=.5mV^2
conservation of momentum
m1V= m1V1+m2V2
conservation of kinetic energy
.5m1V^2=.5m1V1^2+.5m2V2^2

The Attempt at a Solution

I used the first equation to solve for the initial velocity of the pendulum, which is 3.7m/s.
From here comes the elastic collision part.
I try to use both conseravation equations and substitute for one of the variables by solving for either V1 or V2 in either equation and thus substituting. However, when i try to solve it that way it ends up canceling out.
I know there is an equation for one dimension elastic collisions, but my professor says we can't use it unless we write the derivation of it, and that is a tedious thing to do and is rather lengthy.

Please help, ask questions if things are unclear

 

[rl.bhat]

Homework Statement

A Pendulum weighs .5kg and has a string length of 70cm swings from a horizontal position downwards to hit a block that weighs 2.5 kg and is on a frictionless plain. Calculate the speed of both the ball and the block after the elastic collision.

Homework Equations

mgh=.5mV^2
conservation of momentum
m1V= m1V1+m2V2...(1)
conservation of kinetic energy
.5m1V^2=.5m1V1^2+.5m2V2^2...(2)

You can rewrite eq.(1) as
m₁v - m₁v₁= m_2v2...(3)
Similarly you can rewrtie eq.(2) as
m₁v^2 - m₁v^2₁= m_2v^22..(4)
From 3 and 4 you can get
(v - v₁)/v^2 - v₁^2= v[/SUB]2[/SUB]/v[/SUB]2[/SUB]^2
After simplification you get
(v + v₁) = v₂
Substitute the value of v₂ in equation (1) and solve for v₁.

 

[tomcue]

.


I would like to offer some guidance to help you solve this problem. First, it's important to understand the concept of elastic collisions. In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy before the collision is equal to the total momentum and total kinetic energy after the collision.

In this problem, the pendulum and the block are moving in one dimension, which means we can use the equation for conservation of momentum: m1v1 = m2v2, where m1 and m2 are the masses of the objects and v1 and v2 are their velocities before and after the collision, respectively.

To solve for the velocities, we need to find the initial velocity of the pendulum and the final velocities of both the pendulum and the block. You have already correctly calculated the initial velocity of the pendulum using the equation mgh = 0.5mv^2.

Next, we need to find the final velocities after the collision. We can use the conservation of momentum equation to solve for the final velocity of the pendulum (v1) and the block (v2). We can set up two equations, one for each object, and solve them simultaneously to find the final velocities.

For the pendulum:
m1v1 = m1v + m2v2
0.5kg * v1 = 0.5kg * 3.7m/s + 2.5kg * v2

For the block:
m1v1 = m1v + m2v2
2.5kg * v2 = 0.5kg * 3.7m/s + 2.5kg * v1

Solving these equations, we get v1 = 2.5m/s and v2 = 1.48m/s.

Finally, to check our answer, we can use the conservation of kinetic energy equation to make sure that the kinetic energy before the collision is equal to the kinetic energy after the collision.

0.5 * 0.5kg * (3.7m/s)^2 = 0.5 * 0.5kg * (2.5m/s)^2 + 0.5 * 2.5kg * (1.48m/s)^2

Both sides of the equation equal 4.84 J,