Elastic Collision?

Question

A 4.0 kg object is moving at 5.0 m/s EAST. It strikes a 6.0 kg that is at rest. The objects have an elastic collision. The velocity of the 4.0 kg object after the collision is at an angle of 30 degrees SOUTH of EAST. The velocity of the 4.0 kg mass after the collision is?

Answer 1

In an elastic collision both kinetic energy and momentum are conserved.

kinetic energy before = kinetic energy afterwards

v = velocity of first object before collision = 5 m/s east

u = velocity of first object after collision

w = velocity of second object after collision

m = mass of first object = 4 kg

n = mass of second object = 6 kg

(1/2)mv^2 = (1/2)mu^2 + (1/2)nw^2

mv^2 = mu^2 + nw^2

4(5^2) = 4u^2 + 6w^2

100 = 4u^2 + 6w^2

50 = 2u^2 + 3w^2

Momemtum must be conserved. Since there is no movement in the N-S direction, this component of the momentum must be 0. Plus the component in E-W must equal that before the collision.

Momentum before = mv

After collision the first object is moving South of East at 30 degrees.

p(N-S) = m*u*cos(30)

p(E-W) = m*u*sin(30)

For the second object if it moves off at an angle A:

q(N-S) = n*w*cos(A)

a(E-W) = n*w*sin(A)

The N-S components must be equal so:

m*u*cos(30) = n*w*cos(A)

wcos(A) = (m*u)cos(30)/(n) = (4)(u)[SQRT(3)/2] / (6)

wcos(A) = (u/3)SQRT(3)

The momentum before must equal the sum of the E-W after:

mv =  m*u*sin(30) + n*w*sin(A)

4(5) = 4u(1/2) + 6wsin(A)

20 = 2u + 6wsin(A)

10 = u + 3wsin(A)

This gives us three equations:

50 = 2u^2 + 3w^2................. w^2 = [50 - 2u^2]/3

wcos(A) = (u/3)SQRT(3) .... cos(A) = (u/w)SQRT(3)/3

10 = u + 3wsin(A) ................ sin(A) = (10 - u)/(3w)

Since we want to find u, we can use the second two and the fact sin^2 + cos^2 =1  to get an equation in just u and w. We can combine this with equation 1 which only uses u and w to eliminate w and get a solution for u.

sin^2(A) + cos^2(A) = 1

[(10 - u)/(3w)]^2 + [(u/w)SQRT(3)/3]^2 = 1

100 - 20u + u^2 + 3u^2 = 9w^2

From the conservation of energy equation:

50 = 2u^2 + 3w^2 or 9w^2 = 150 - 6u^2

100 - 20u + u^2 + 3u^2 = 150 - 6u^2

10u^2 - 20u - 50 = 0

u^2 - 2u - 5 = 0

u = [2 +/- SQRT(4 + 20] / 2

u = 1 +/- SQRT(6)

u = 3.45 m/s

Answer 2

I don't think so...

You don't need all that...PLUS there is a math error in it...

The momentum equation in the EAST-WEST direction should be COSINE Not SINE.  The answer will thus be wrong.

I could have an error somewhere, too...it is easy to make aritmatic errors when doing these.  If you follow my solution, you will not need to solve a quadratic either.  Use the method you find easiest to do on your own.  The physics is the same.  The order in which we did things is a bit different.

1) Conserve momentum (in the vertical, i.e., North-South) and horizontal (i.e., East-West) directions...

Call x the angle that the 6 kg mass makes with the East-West axis.

NoSo dir: (Conservation of momentum)...

4 Vsin(30) = 6vsin(x)  where V  velocity of 4 kg mass after collision and v is the 6 kg mass's velocity. And simplifying...

(1/3) V = vsin(x)

EaWe dir: (Conservation of momentum)...

4 Vcos (30) = 6vcos(x)...so we have

(1/3) V cos(30) = vcos(x) and so,

(1/3)Sqrt(3)V = vcos(x)

Now divide  (1/3) V = vsin(x) by (1/3)Sqrt(3)V = vcos(x) and get..

tan(x) = SQRT(3) and x=60 degrees

[Note: When we have a 2D system in an ellastic collision, the two masses will form a right angle when energy is conserved]

The energy at the start is 1/2 m 5^2 = 50 J and after it is also 50J

Take (1/3) V = vsin(x) with x=60 to get v=2/9 SQRT(3) V.  The Kinetic energy after is...

1/2 4 V^2 + 1/2 6 [(2/9) * SQRT(3) V]^2 or

2 V^2 + (4/9) V^2 = 50J so that V=4.52 m/s

The other mass has a velocity of 1.74 m/s

-Fred