Question

A 30 g of clay travelling east at 3.0 m/s collides with a 20 g ball of clay travelling north at 4.0 m/s. What are the speed and the direction of resulting 50 g blob of clay?

Answer 1

This is a simple "conservation of momentum" problem.

Total momentum before collision = total momentum after collision

In this problem, direction is important; so we have to use vector math.

To express that equation, start by defining some variables:

m₁, m₂ = masses of the balls of clay (30g, 20g);
V₁, V₂ = initial velocity VECTORS of the balls;
U = final velocity VECTOR of the combined blob.

Then the conservation equation becomes:

m₁V₁ + m₂V₂ = (m₁+m₂)U

This breaks down into two component equations; one for the x-direction and the other for the y-direction:

m₁Vx₁ + m₂Vx₂ = (m₁+m₂)Ux
m₁Vy₁ + m₂Vy₂ = (m₁+m₂)Uy

We're interested in the value of "U" (the final velocity vector), so rewrite the above two equations like this:

Ux = (m₁Vx₁+m₂Vx₂)/(m₁+m₂)
Uy = (m₁Vy₁+m₂Vy₂)/(m₁+m₂)

Now, if we take the positive x-direction to be "east" and the positive y-direction to be "north", then from the conditions given in the problem we have:

Vx₁ = 3.0 meters/s; Vy₁ = 0
Vx₂ = 0; Vy₂ = 4 meters/s

Now plug those values into the Ux, Uy equations; and also plug in the given values for m₁, m₁:

Ux = (30g*3.0 meters/s) / 50g
Uy = (20g*4 meters/s) / 50g

(Now push the buttons on your calculator).

Once you have Ux, Uy, you can answer the questions:

1.What is the movement direction of the resulting 50g blob of clay?

It moves at an angle θ relative to the x-axis, given by:

tanθ = Uy/Ux
(solve for "θ". Express this in terms of "north" and "east", like: so many degrees north of east)

2.What is the speed of the resulting 50g blob of clay?

By the Pythagorean theorem:

speed = √(Ux²+Uy²= sqrt((30*3)/50)^2+((20*4)/50)^2=2.40 m/s