Two–Dimensional Collisions Problems and Solutions

Problem #1
A 50.0­kg skater is travelling due east at a speed of 3.00 m/s. A 70.0­kg skater is moving due south at a speed of 7.00 m/s. They collide and hold on to one another after the collision, managing to move off at an angle θ south of east with a speed v_f . Find (a) the angle θ and (b) the speed v_f , assuming that friction can be ignored.

Fig.1

Answer:
In any kind of collision, momentum is conserved so

(m₁ + m₂)v_f = m₁v_1i + m₂v_2i .                         (1)

Now momentum and velocity are vector quantities and the i and j components must be handled separately

(m₁ + m₂)v_fx = m₁v_1ix + m₂v_2ix ,                     (1a)
(m₁ + m₂)v_fy = m₁v_1iy + m₂v_2iy .                    (1b)

So we can rearrange these equations to find the components of the final velocity

v_fx = (m₁v_1ix + m₂v_2ix)/(m₁ + m₂)                   (2a)
v_fy = (m₁v_1iy + m₂v_2iy)/(m₁ + m₂)                                  (2b)

Using the given values, we find

v_fx = [(50 kg)(3 m/s) + (70 kg)(0)]/(50 kg + 70kg) = 1.25 m/s           (2a)
v_fy = [(50 kg)(0) + (70 kg)(­7 m/s)/(50 kg + 70 kg) = 4.083 m/s         (2b)

To find the magnitude and direction of the final velocity, we use the Pythagorean Theorem and trigonometry

v_f = [(v_fx)² + (v_fx)²]^1/2 = 4.27 m/s, and
θ = arctan|v_fy/v_fx| = 72.98°.

The final velocity of the pair is 4.27 m/s at 73.0° south of east.

Problem #2
Two opposing hockey players are racing up the ice for the puck when they collide at point A as shown in the diagram below. The first hockey player has mass 90 kg and a speed of 2.7 m/s while the other has mass 82 kg and speed 3.1 m/s. The angle in the diagram is θ = 32°. After the collision, the players remain locked together (at least until the referee forces them apart). What is the magnitude and direction of the players' velocity just after they collide?

Fig.2

Answer:
Since the collision is totally inelastic and in two dimensions, we find that we are dealing with a vector addition problem, P_T = P₁ + P₂. First we calculate the magnitude of each player's momentum using p = mv,

P₁ = m₁v₁ = (90 kg)(2.7 m/s) = 243 kg­m/s ,
P₂ = m₂v₂ = (82 kg)(3.1 m/s) = 254.2 kg­m/s.

Then we find P_T by the component method,
momentum on the x axis
P_1x = 0
P_2x = 254.2 kg.m/s sin(32°) = = 134.705 kg.m/s
P_Tx = 134.705 kg.m/s

momentum on the y axis
P_1y = 243 kg.m/s
P_2y = 254.2 kg.m/s cos(32°) = 215.574 kg.m/s
P_Ty = 458.574 kg.m/s

Using the Pythagorean formula we find,

P_T = [(P_Tx)² + (P_Ty)²]^1/2 = [(134.705 kg.m/s)² + (458.574 kg.m/s)²]^1/2 = 477.949 kg.­m/s .

Using trigonometry, we find the angle from

θ = arctan(P_Ty/P_Tx) = arc tan(458.574/134.705) = 73.63°.

So the total momentum of the two players is P_T = (478,73.6°).

Now P_T = (m₁ + m₂)v_f , so the final velocity must be in the same direction as the total momentum. The magnitude of the velocity is

v_f = P_T/(m₁ + m₂) = 477.949 kg­m/s/ (90 kg + 82 kg) = 2.78 m/s

So the final velocity of the two players just after the collision is v_f = (2.78 m/s, 73.6°).

Problem #3
An unstable nucleus of mass 17 × 10⁻²⁷ kg initially at rest disintegrates into three particles. One of the particles, of mass 5.0×10⁻²⁷ kg, moves along the y axis with a speed of 6.0 × 10⁶ m/s. Another particle of mass 8.4 × 10⁻²⁷ kg, moves along the x axis with a speed of 4.0 × 10⁶ m/s . Find (a) the velocity of the third particle and (b) the total energy given off in the process.

Fig.3

Answer:
(a) First, draw a picture of what is happening! Such a picture is given in Fig. 3. In the most general sense of the word, this is indeed a “collision”, since it involves the rapid interaction of a few isolated particles.
The are no external forces acting on the particles involved in the disintegration; the total momentum of the system is conserved. The parent nucleus is at rest , so that the total momentum was (and remains) zero: Pi = 0.
Afterwards, the system consists of three particles; for two of these particles, we are given the masses and velocities. We are not given the mass of the third piece, but since we were given the mass of the parent nucleus, we might think that we can use the fact that the masses must sum up to the same value before and after the reaction to find it. In fact relativity tells us that masses don’t really add in this way and when nuclei break up there is a measurable mass difference, but it is small enough that we can safely ignore it in this problem. So we would say that mass is conserved, and if m is the mass of the unknown fragment, we get:

17 × 10⁻²⁷ kg = 5.0 × 10⁻²⁷ kg + 8.4 × 10⁻²⁷ kg + m

so that

m = 3.6 × 10⁻²⁷ kg .

We will let the velocity components of the third fragment be vx and v_y. Then the total x momentum after the collision is

P_f,x = (8.4 × 10⁻²⁷ kg)(4.0 × 10⁶ m/s) + (3.6 × 10⁻²⁷ kg)v_x

Using P_i,x = 0 = P_f,x, we find:

(8.4 × 10⁻²⁷ kg)(4.0 × 10⁶ m/s) + (3.6 × 10⁻²⁷ kg)v_x = 0
which easily gives:

v_x = −9.33 × 10⁶ m/s

Similarly, the total y momentum after the collision is:

P_f,y = (5.0 × 10⁻²⁷ kg)(6.0 × 10⁶ m/s) + (3.6 × 10⁻²⁷ kg)v_y

and using P_i,y = 0 = P_f,y, we have:

(5.0 × 10⁻²⁷ kg)(6.0 × 10⁶ m/s) + (3.6 × 10⁻²⁷ kg)v_y = 0

which gives

v_y = −8.33 × 10⁶ m/s

This really does specify the velocity of the third fragment (as requested), but it is also useful to express it as a magnitude and direction. The speed of the third fragment is

v = [(−9.33 × 10⁶ m/s)² + (−8.33 × 10⁶ m/s)²]^1/2 = 1.25 × 10⁷ m/s

and its direction θ (measured counterclockwise from the x axis) is given by

tan θ = [−8.33/−9.33] = 0.893

Realizing that θ must lie in the third quadrant, we find:

θ = tan⁻¹(0.893) – 180⁰ = −138⁰ .

(b) What is the gain in energy by the system for this disintegration? By this we mean the gain in kinetic energy. Initially, the system has no kinetic energy. After the breakup, the kinetic energy is the sum of ½ mv² for all the particles, namely

K_f = ½ (5.0 × 10⁻²⁷ kg)(6.0 × 10⁶ m/s)² + ½ (8.4 × 10⁻²⁷ kg)(4.0 × 10⁶ m/s)² + ½ (3.6 × 10⁻²⁷ kg)(1.25 × 10⁷ m/s)² =  4.38 × 10⁻¹³ J

We might say that the process gives off 4.38 × 10⁻¹³ J of energy.

Problem #4
A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30.0⁰ with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball’s velocity.

Fig.4

Answer:
The collision is diagrammed in Fig. 4. We don’t know the final speed of the struck ball; we will call it v, as in Fig. 4(b). We don’t know the final direction of motion of the struck ball; we will let it be some angle θ, measured below the x axis, also as shown in Fig. 7.13(b).
Since we are dealing with a “collision” between the two objects, we know that the total momentum of the system is conserved. So the x and y components of the total momentum is the same before and after the collision.
Suppose we let the x and y components of the struck ball’s final velocity be vx and vy, respectively. Then the condition that the total x momentum be conserved gives us:

m(5.00 m/s) + 0 = m(4.33 m/s) cos 30.0⁰ + mv_x

(The struck ball has no momentum initially; after the collision, the incident ball has an x velocity of (4.33 m/s) cos 30.0⁰.) Luckily, the m’s cancel out of this equation and we can solve for v_x:

(5.00 m/s) = (4.33 m/s) cos 30.0⁰ + v_x

and then:

v_x = (5.00 m/s) − (4.33 m/s) cos 30.0⁰ = 1.25 m/s

We can similarly find vy by using the condition that the total y momentum be conserved in the collision. This gives us:

0 + 0 = m(4.33 m/s) sin 30.0⁰ + mv_y

which gives

v_y = −(4.33 m/s) sin 30.0⁰ = −2.16 m/s

Now that we have the components of the final velocity we can find the speed and direction of motion. The speed is:

v = [v²_x + v²_y]^1/2 = [(1.25 m/s)² + (−2.16 m/s)²]^1/2 = 2.50 m/s

and the direction is found from
tan θ = [v_y/v_x] = [−2.16 m/s/1.25 m/s] = −1.73

θ = tan⁻¹(−1.73) = −60⁰

So the struck ball moves off with a speed of 2.50 m/s at an angle of 60⁰ downward from the x axis.
This really completes the problem but we notice something strange here: We were given more information about the collision than we used. We were also told that the collision was elastic, meaning that the total kinetic energy of the system was the same before and after the collision. Since we now have all of the speeds, we can check this.
The total kinetic energy before the collision was

K_i = ½ m(5.00 m/s)² = m(12.5 m²/s²)

The total kinetic energy after the collision was

K_f = ½ m(4.33 m/s)² + ½ m(2.50 m/s)² = m(12.5 m²/s²)

so that K_i is the same as K_f ; the statement that the collision is elastic is consistent with the other data; but in this case we were given too much information in the problem!