X: m1v1i cos θi+ m2v2i cos φi = m1v1f cos θf + m2v2f cosφi
Y: – m1v1i sin θi+ m2v2i sin φi = m1v1f sinθf – m2v2f sinφi
Q1) In the game of billiards, all the balls have
approximately the same mass, about 0.17kg. In the figure,
the cue ball strikes another ball at rest such that it follows
the path shown. The other ball has a speed of 1.5m/s
immediately after the collision. What is the speed of the
cue ball after the collision?
a) 2.1m/s
b) 2.6m/s
c) 3.3m/s
d) 4.9m/s
3m/s
30°
60°
1.5m/s
Sol) The cue ball(m1) was moving only horizontally before collision, so v1ix= v1i cos0° = v1i and v1iy = 0.
The other ball(m2) was stationary, so v2ix = 0 and v2iy = 0
X : m1v1i = m1v1f cos30° + m2v2f cos(-60°)
Y : 0 = m1v1f sin30° + m2v2f sin(-60°)
The mass are identical, so they cancel out. And inserting v1i =3m/s and v2f =1.5m/s gives
X : 3 = v1f cos30° + 1.5× cos60°
Y : 0 = v1f sin30° – 1.5× sin60°
We can use either equation above to solve v1f but let's use the Y-component equation since it is more
simple,
0 = v1f sin30° – 1.5× sin60°
=> v1f sin30° =1.5× sin60°
=> v1f =
= 2.6m/s
X: m1v1i cos θi+ m2v2i cos φi = m1v1f cos θf + m2v2f cosφi
Y: – m1v1i sin θi+ m2v2i sin φi = m1v1f sinθf – m2v2f sinφi
Q2) A 1500kg car traveling east with a speed of 25m/s
collides at an intersection with a 2500kg van traveling
north at a speed of 20m/s. After collision, the two cars
stick together for brief period of time. Find the
direction and the speed of the ‘wreckage’ after
collision.
a) φ=53.1°, 15.6m/s
b) φ=53.1°, 11.9m/s
c) φ=28.4°, 15.6m/s
d) φ=28.4°, 11.9m/s
vf
φ=?
25m/s
20m/s
Sol) mass of the car m1=1500kg, speed before the collision v1i =25m/s
mass of the van m2=2500kg, speed before the collision v2i =20m/s
The car was moving only horizontally before the collision, so v1ix= v1i cos0° = v1i and v1iy = 0.
The van was moving only vertically before the collision , so v2iy= v1i sin90° = v2i and v2ix = 0.
And since the two vehicles combine after collision, we set the mass after collision as (m1 + m2);
X : m1v1i = (m1 + m2 )vf cosφ
Y: m2v2i = (m1 + m2 )vf sinφ
So
X : 1500×25 = (1500+2500 )vf cosφ -------------(1)
Y: 2500×20 =(1500+2500 )vf sinφ -------------(2)
If we do (2)÷(1), then
1.33 = tan φ
=> φ = tan-1(1.33) = 53.1°
To find the speed vf , we simply use the now know angle and plug it into (1) or (2). So vf =15.6m/s
X: m1v1i cos θi+ m2v2i cos φi = m1v1f cos θf + m2v2f cosφi
Y: – m1v1i sin θi+ m2v2i sin φi = m1v1f sinθf – m2v2f sinφi
Q3) A 0.5kg bomb is sliding along an icy pond(frictionless) with a velocity of 2m/s to the right. The
bomb explodes into two pieces. After the explosion, a 0.2kg piece moves south at 4m/s. What is the speed
of the 0.3kg piece and what angle(φ) does it make in respect to the x-axis?
a) 2.9m/s, φ=25°
b) 2.9m/s, φ=39°
c) 4.3m/s, φ=25°
d) 4.3m/s, φ=39°
φ=?
y
x
Sol) Knowing that the bomb explodes into two pieces m 1 and m2, where m1=0.3kg and m2=0.2kg, we
can say the mass of the bomb before the explosion is (m1 + m2)
The bomb is moving only horizontally before collision, and since m2 moves vertically downward after
collision,
X : (m1 + m2 )v1i = m1v1f cosφ
Y : 0 = m1v1f sinφ – m2v2f
Inserting m1=0.3kg, m2=0.2kg, v1i =2m/s, v2f =4m/s, we obtain
X : 0.5×2 = 0.3×v1f cosφ
Y : 0 = 0.3×v1f sinφ – 0.2×4
And then after some rearranging,
0.3×v1f cosφ=1 ----------------(1)
0.3×v1f sinφ = 0.8--------------(2)
If we do (2)÷(1),
tanφ=0.8
so
φ=tan-1(0.8)=38.7°
Knowing the angle φ, we can now solve for v1f by inserting φ=38.7°into (1) or (2) above, which is
v1f = 4.27m/s
X: m1v1i cos θi+ m2v2i cos φi = m1v1f cos θf + m2v2f cosφi
Y: – m1v1i sin θi+ m2v2i sin φi = m1v1f sinθf – m2v2f sinφi
φ=?
Q4) Two asteroids are drifting in space with
trajectories shown. The mass of m2 is ten times
more massive than m1. After the collision they stick
together briefly. If the speed of m1 and m2 before
collision is 45m/s and 15m/s respectively, at what
angle do the combined asteroids deflect from the
x-axis?
θ=43°
m1
m2
Sol)
X : m1v1i cos43° = (m1 + m2 )vf cosφ
Y: m1v1i sin43° + m2v2i = (m1 + m2 )vf sinφ
Inserting v1i =45m/s, v2i =15m/s and since m2 is ten times larger than m1, we can set m2=10m and
m1=m. So we can obtain
X : m×45×cos43° = (m + 10m )vf cosφ
Y: m×45×sin43° + 10m ×15 = (m + 10m )vf sinφ
Cancel out 'm' and after some modification
X : 32.9 = 11vf cosφ ---------(1)
Y: 180.7 = 11vf sinφ --------(2)
If we (2)÷(1)
tanφ=5.49
so
φ=tan-1(5.49)=80°