2-D collison question involving one moving objecy and one stationary object

[meeklobraca]

Homework Statement

A 20.0kg curling rock is moving due east when it collides with a stationary rock of equal mass. After the collison the first rock is traveling at 1.50 m/s 47 degrees north of east, and the other rock is traveling 1.71 m/s 40 degrees south of east. Determine the velocity of the first rock before the collison.

Homework Equations

p=mv

The Attempt at a Solution

I can sort of figure out how to do this if your given the velocity of the moving rock, but I am kind of at a loss as to how to figure it out going the other way. THe attempt I made was that the momentum before and after have to equal the same. So I figured the momentum of the two rocks after the collison equalled 64.2N. but taking that and diving it by the mass of the before rock doesn't equal the answer.

Can someone point me in the right direction for trying to figure out this question?

Thanks!

 

[anathema]

Hello,

To solve this problem, you can use the conservation of momentum principle, which states that the total momentum of a system before and after a collision remains constant. This can be written as:

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 and m2 are the masses of the two rocks, v1i and v2i are their initial velocities before the collision, and v1f and v2f are their final velocities after the collision.

In this problem, we are given the masses of the rocks (20.0kg each) and the final velocities (1.50 m/s 47 degrees north of east and 1.71 m/s 40 degrees south of east). We can use this information to solve for the initial velocity of the first rock (v1i).

First, we need to break down the final velocities into their x and y components. To do this, we can use the fact that the direction of the velocity vector is given by the angle (47 degrees north of east or 40 degrees south of east) and the magnitude of the velocity vector is given by the speed (1.50 m/s or 1.71 m/s).

For the first rock, we can write its final velocity as:

v1f = 1.50 m/s x cos(47 degrees) + 1.50 m/s x sin(47 degrees)

= 1.14 m/s + 0.96 m/s

= 2.10 m/s

Similarly, for the second rock, we can write its final velocity as:

v2f = 1.71 m/s x cos(40 degrees) - 1.71 m/s x sin(40 degrees)

= 1.31 m/s - 1.14 m/s

= 0.17 m/s

Now, we can substitute these values into the conservation of momentum equation and solve for v1i:

m1v1i + m2v2i = m1v1f + m2v2f

(20.0kg)(v1i) + (20.0kg)(0 m/s) = (20.0kg)(2.10 m/s) + (20.0kg)(0.17 m/s)

20.0kg x v1i =

 

[baba89]

I would approach this problem by first defining the variables and determining what information is given and what is needed. In this case, the given information includes the masses of the two rocks, the velocity of the first rock after the collision, and the angles at which the two rocks are travelling after the collision. The needed information is the velocity of the first rock before the collision.

Next, I would use the conservation of momentum principle to solve for the velocity of the first rock before the collision. This principle states that the total momentum before and after a collision must be equal. So, we can set up the equation:

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 and m2 are the masses of the two rocks, v1i and v2i are the velocities of the two rocks before the collision, and v1f and v2f are the velocities of the two rocks after the collision.

Since the second rock is initially stationary (v2i = 0), the equation simplifies to:

m1v1i = m1v1f + m2v2f

Substituting in the given values, we get:

20.0kg * v1i = 20.0kg * (1.50 m/s * cos47°) + 20.0kg * (1.71 m/s * cos40°)

Solving for v1i, we get:

v1i = (1.50 m/s * cos47°) + (1.71 m/s * cos40°)

v1i = 1.05 m/s + 1.17 m/s

v1i = 2.22 m/s

So, the velocity of the first rock before the collision was 2.22 m/s. This approach uses the conservation of momentum principle and trigonometric functions to solve for the unknown velocity.