2 Dimensional Collision Problem

These are not the appropriate equations for the problem.
It is a two-dimensional collision, both components of the momentum has to be conserved. Make a drawing and write the equations for the components, both parallel and perpendicular to the initial velocity of the 1.9 kg object. There are enough data, you do not need conservation of energy.

ehild

 

Do you know how to get the x and y components of a vector if you know the magnitude and the angle the vector encloses with the x axis?
Do you know what momentum is? What does conservation of momentum mean?

Show something you did to solve this problem. I will tell you if it is correct or not.

ehild

 

I see... I seriously suggest that you read your book or notes about vectors.

Velocity is a vector. It has magnitude and direction. A vector can be represented in a Cartesian coordinate system by an arrow, inclined with an angle with respect to the x axis. Its components are projections on the axes, see figure. The red vector, [itex]\vec a[/itex] has components a_x and a_y, the components of the blue vector [itex]\vec b[/itex] are b_x and b_y.
When you add vectors, their x and y components add separately. The resultant vector [itex]\vec c=\vec a +\vec b[/itex] has the components c_x=a_x+b_x and cy=a_y+b_y.

Never add x component to y component!!

The momentum is vector, too, [itex]\vec p=m \vec v[/itex] Its components are
p_x=mv_x and p_y=mv_y.

The total momentum of two objects is the sum of the individual momenta. [itex]\vec p=\vec p_1+\vec p_2[/itex]. That means two equations for the components:

p_x=p_1x+p_2x and p_y=p_1y+p_2y.

Conservation of momentum during a collision means that the total momentum before the collision is the same as the total momentum after the collision.

[itex]\vec p(initial)=\vec p(final)[/itex]

That is valid to the x and y components separately.

Now we start with this problem, writing out the individual components of the initial and final momenta for both objects.

The mass of object 1 is m₁=1.9 kg,, the mass of object 2 is m₂=1.4 kg.

We set up a coordinate system with x axis in the direction of the velocity of the first object.

Initially the components of momentum of the first object are

p_1x(i)=m₁v_1x(i)=1.9*4.9 cos(0)=1.9*4.9
p_1y(i)=m₁v_1y(i)=1.9*4.9 sin(0)=0

Te momentum of the second object is zero:
p_2x(i)=0
p_2y(i)=0.

After the collision, the components of momentum for the first object:

p_1x(f)=1.9*1.2 cos(41)
p_1y(f)=1.9*1.2 sin(41)

Those for the second object (with speed v₂(f) and direction of velocity θ)

p_2x(f)=1.4*v₂(f) cos(θ)
p_2y(f)=1.4*v₂(f) sin(θ)

Apply conservation of momentum for both the x and y components separately.

p_1x(i)+p_2x(i)=p_1x(f)+p_2x(f)

1.9*4.9 =1.9*1.2 cos(41)+1.4*v₂(f) cos(θ)

0=1.9*1.2 sin(41)+1.4*v₂(f) sin(θ)

these are two equations for two unknowns: the speed of object 2, v₂(f), and the angle of its velocity, θ. You can solve.

ehild