Two-Dimensional Collisions Problems and Solutions
Problem#1
A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. If the collision is perfectly inelastic, (a) calculate the speed and direction of the players just after the tackle and (b) determine the mechanical energy lost as a
result of the collision. Account for the missing energy.
Answer:
m1v1i + m2v2i = (m1 + m2)v
for the system of two football players in the x direction (the direction of travel of the fullback).
m1v1i + m2v2i = m1v1f + m2v2f
(90.0 kg)(5.00 m/s) + 0 = (90.0 kg + 95.0 kg)V cosθ
where θ is the angle between the direction of the final velocity V and the x axis. We find
450kgm/s = (185kg)V cosθ
V cosθ = 2.43 m/s (1)
Now consider conservation of momentum of the system in the y direction (the direction of travel of the opponent).
(95.0 kg)(3.00 m/s) + 0 = (90.0 kg + 95.0 kg)V sinθ
285 kgm/s = 185kgVsinθ
Vsinθ = 1.54 m/s (2)
Divide equation (2) by (1)
tanθ = 0.633
then, θ = tan-1(0.633) = 32.30
Then, either (1) or (2) gives
V cos32.30 = 2.43 m/s
V = 2.88 m/s
(b) initial kinetic energy of the system is
Ki = ½ m1v1i2 + ½ m2v2i2
Ki = ½ (90.0 kg)(5.00 m/s)2 + ½ (95.0 kg)(3.00 m/s)2 = 1.55 kJ
final kinetic energy of the system is
Kf = ½ (m1 + m2)V2
Kf = ½ (90.0 kg + 95.0 kg)(2.88 m/s)2 = 0.767 kJ
Thus, the kinetic energy lost is
∆K = 1.55 kJ – 0.767 kJ = 0.783 kJ
Problem#2
Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed of 5.00 m/s. After the collision, the orange disk moves along a direction that makes an angle of 37.0° with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.
Answer:
The law of conservation of the momentum of the collision on the x-axis
pxi = pxf
mvOi + mvYi = mvOf + mvYf
(5.00 m/s) + 0 = vOf cos37.00 + vYfcos53.00
0.799vOf + 0.602vYf = 5.00 (1)
The law of conservation of the momentum of the collision on the y-axis
0 + 0 = vOf sin37.00 – vYfsin53.00
(0.602)vOf = (0.799)vYf
Solving (1) and (2) simultaneously,
0.799vOf + 0.602(0.602vOf/0.799) = 5.00
1.55vOf = 5.00
vOf = 3.99 m/s and vYf = 3.01 m/s
Problem#3
Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed vi. After the collision, the orange disk moves along a direction that makes an angle & with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk.
Answer:
The law of conservation of the momentum of the collision on the x-axis
pxi = pxf
mvOi + mvYi = mvOf + mvYf
vi = v cosθ + ucos(90.00 – θ)
vi = v cosθ + usinθ
The law of conservation of the momentum of the collision on the y-axis
0 + 0 = v sinθ – usin(90.00 – θ)
v sinθ = ucosθ
Solving (1) and (2) simultaneously,
vi = (ucosθ/sinθ)cosθ + usinθ
vi sinθ = ucos2θ + usin2θ = u(cos2θ + sin2θ)
u = vi sinθ
and from (2) we get
v = vi cosθ
We did not need to write down an equation expressing conservation of mechanical energy. In the
problem situation, the requirement of perpendicular final velocities is equivalent to the condition of
elasticity.
Problem#4
The mass of the blue puck in Figure 4 is 20.0% greater than the mass of the green one. Before colliding, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds of the pucks after the collision if half the kinetic energy is lost during the collision.
Answer:
The collision is inelastic, since energy is not conserved. The total momentum of the two pucks is zero before the collision and after the collision.
mGvGi + mBvBi = 0
m(10.0 m/s) – (1.20m)vBi = 0
vBi = 8.33 nm/s
and
mGvGf + mBvBf = 0
mvGf – (1.20 m)vBf = 0
vGf = 1.20vBf
The final kinetic energy of the system equals ½ times its initial kinetic energy, then
Kf = ½ Ki
½mGvGi2 + ½mBvBi2 = 2(½mGvGf2 + ½mBvBf2)
m(10.0 m/s)2 + 1.20m(8.33 m/s)2 = 2mvGf2 + 2(1.20m)vBf2
vGf2 + 1.20vBf2 = 91.7 m2/s2 (2)
Solving (1) and (2) simultaneously, we find
(1.20vBf)2 + 1.20vBf2 = 91.7 m2/s2
2.64vBf2 = 91.7 m2/s2
vBf = 5.89 m/s
then from (1).
vGf = 1.20vBf = 7.07 m/s
Problem#5
An object of mass 3.00 kg, moving with an initial velocity of 5.00i m/s, collides with and sticks to an object of mass 2.00 kg with an initial velocity of -3.00j m/s. Find the final velocity of the composite object.
Answer:
We use the law of conservation of momentum, pi = pf, then
m1v1i + m2v2i = (m1 + m2)vf
(3.00 kg)(5.00i m/s) + (2.00 kg)(–3j m/s) = 5.00v
v = (3.00i – 1.20j) m/s
pxi = pxf
mvOi + mvYi = mvOf + mvYf
vi = v cosθ + ucos(90.00 – θ)
vi = v cosθ + usinθ
The law of conservation of the momentum of the collision on the y-axis
0 + 0 = v sinθ – usin(90.00 – θ)
v sinθ = ucosθ
Solving (1) and (2) simultaneously,
vi = (ucosθ/sinθ)cosθ + usinθ
vi sinθ = ucos2θ + usin2θ = u(cos2θ + sin2θ)
u = vi sinθ
and from (2) we get
v = vi cosθ
We did not need to write down an equation expressing conservation of mechanical energy. In the
problem situation, the requirement of perpendicular final velocities is equivalent to the condition of
elasticity.
Problem#4
The mass of the blue puck in Figure 4 is 20.0% greater than the mass of the green one. Before colliding, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 10.0 m/s. Find the speeds of the pucks after the collision if half the kinetic energy is lost during the collision.
Answer:
The collision is inelastic, since energy is not conserved. The total momentum of the two pucks is zero before the collision and after the collision.
mGvGi + mBvBi = 0
m(10.0 m/s) – (1.20m)vBi = 0
vBi = 8.33 nm/s
and
mGvGf + mBvBf = 0
mvGf – (1.20 m)vBf = 0
vGf = 1.20vBf
The final kinetic energy of the system equals ½ times its initial kinetic energy, then
Kf = ½ Ki
½mGvGi2 + ½mBvBi2 = 2(½mGvGf2 + ½mBvBf2)
m(10.0 m/s)2 + 1.20m(8.33 m/s)2 = 2mvGf2 + 2(1.20m)vBf2
vGf2 + 1.20vBf2 = 91.7 m2/s2 (2)
Solving (1) and (2) simultaneously, we find
(1.20vBf)2 + 1.20vBf2 = 91.7 m2/s2
2.64vBf2 = 91.7 m2/s2
vBf = 5.89 m/s
then from (1).
vGf = 1.20vBf = 7.07 m/s
Problem#5
An object of mass 3.00 kg, moving with an initial velocity of 5.00i m/s, collides with and sticks to an object of mass 2.00 kg with an initial velocity of -3.00j m/s. Find the final velocity of the composite object.
Answer:
We use the law of conservation of momentum, pi = pf, then
m1v1i + m2v2i = (m1 + m2)vf
(3.00 kg)(5.00i m/s) + (2.00 kg)(–3j m/s) = 5.00v
v = (3.00i – 1.20j) m/s
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