PHY 2053, Spring 2009, Quiz 7 — Whiting
1. A 2,000-kg car moving East at 10.0 m/s collides with a 3,000-kg car moving North. After
the collision, the cars stick together and move as a unit, at an angle of 60◦ North of East.
a) Find the speed of the cars after the collision.
East-momentum: m1 × v1 = (m1 + m2 ) × vf cos(60◦ ). Thus:
vf =
m1 × v1
2000 × 10.0
=
= 8.00 m/s.
◦
(m1 + m2 ) × cos(60 )
(2000 + 3000) × 0.5
b) Find the speed of the 3,000-kg car before the collision.
North-momentum: m2 × v2 = (m1 + m2 ) × vf sin(60◦ ). Thus:
v2 =
(m1 + m2 ) × vf sin(60◦ )
(2000 + 3000) × 8.00 × 0.8660
=
= 11.55 m/s.
m2
3000
Check: should have m2 × v2 = m1 × v1 tan(60◦ ) by trigonometry, using momenta as vectors.
2. A 10.0-g object moving to the right at 20.0 cm/s makes an elastic head-on collision with
a 5.0-g object that is initially at rest.
a) Find the velocity of each object after the collision.
Momentum: m1 × v1i = m1 × v1f + m2 × v2f .
Elastic: (v1i − v2i ) = −(v1f − v2f )
⇒
v2f = v1i + v1f .
Thus: m1 × v1i = m1 × v1f + m2 × (v1i + v1f ). We can solve or v1f :
v1f =
(10.0 − 5.0) × 20.0
(m1 − m2 ) × v1i
=
= 6.67 cm/s; v2f = 20.0 + 6.67 = 26.67 cm/s.
(m1 + m2 )
(10.0 + 5.0)
Check: 10.0 × 20.0 = 10.0 × 6.67 + 5.0 × 26.67.
b) Find the fraction of the initial kinetic energy transferred to the 5.0-g object.
2 . KE = 1 m × v 2 . KE = 1 m × v 2 .
KEi = 21 m1 × v1i
2f
1f
2 2
2 1
2f
1f
Ratio =
2
m2 × v2f
5.0 × (26.67)2
=
= 88.9%.
2
10.0 × (20.0)2
m1 × v1i
Check:
Balance =
2
m1 × v1f
(6.67)2
=
= 11.1%.
2
(20.0)2
m1 × v1i