Use the worked example above to help you solve this problem. A car with mass 1.55 x 10³ kg traveling east at a speed of 20.8 m/s collides at an intersection with a 2.55 x 10³ kg van traveling north at a speed of 19.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. m/s ° counterclockwise from the +x-axis magnitude direction

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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GOAL Analyze a two-dimensional inelastic collision.
PROBLEM A car with mass 1.50 x 10³ kg traveling east at a
speed of 25.0 m/s collides at an intersection with a 2.50 x 10³
kg van traveling north at a speed of 20.0 m/s, as shown in the
figure. Find the magnitude and direction of the velocity of the
wreckage after the collision, assuming that the vehicles
undergo a perfectly inelastic collision (that is, they stick
together) and assuming that friction between the vehicles and
the road can be neglected.
SOLUTION
Find the x-components of the initial
and final total momenta.
STRATEGY Use conservation of momentum in two
dimensions. (Kinetic energy is not conserved.) Choose
coordinates as in the figure. Before the collision, the only
object having momentum in the x-direction is the car, while
the van carries all the momentum in the y-direction. After the totally inelastic collision, both vehicles move
together at some common speed vf and angle 0. Solve for these two unknowns, using the two components
of the conservation of momentum equation.
A top view of a perfectly inelastic collision between a car
and a van.
Set the initial x-momentum equal to
the final x-momentum.
Find the y-components of the initial
and final total momenta.
Set the initial y-momentum equal to
the final y-momentum.
Divide Equation (2) by Equation (1)
and solve for 0.
Σpxi
mcarV car
= 3.75 x 104 kg. m/s
Σpxf = (mcar + mvan) vf cos 0 = (4.00 × 10³ kg) vf cos (
ΣΡίγ
(1) 3.75 x 104 kg
tan 0
=
Ꮎ
+25.0 m/s
=
=
(1.50 x 10³ kg)(25.0 m/s)
mvan van
= 5.00 x 104 kg. m/s
=
Σpfy = (mcar + mvan) vf sin 0 (4.00 x 10³ kg) vf sin 0
(2) 5.00 x 104 kg · m/s = (4.00 × 10³ kg) vf sin 0
53.1°
x
+20.0 m/s
m/s = (4.00 × 10³ kg) vf cos 0
(2.50 x 10³ kg)(20.0 m/s)
5.00 x 104 kg. m/s
3.75 x 104 kg. m/s
= 1.33
Transcribed Image Text:GOAL Analyze a two-dimensional inelastic collision. PROBLEM A car with mass 1.50 x 10³ kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 x 10³ kg van traveling north at a speed of 20.0 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. SOLUTION Find the x-components of the initial and final total momenta. STRATEGY Use conservation of momentum in two dimensions. (Kinetic energy is not conserved.) Choose coordinates as in the figure. Before the collision, the only object having momentum in the x-direction is the car, while the van carries all the momentum in the y-direction. After the totally inelastic collision, both vehicles move together at some common speed vf and angle 0. Solve for these two unknowns, using the two components of the conservation of momentum equation. A top view of a perfectly inelastic collision between a car and a van. Set the initial x-momentum equal to the final x-momentum. Find the y-components of the initial and final total momenta. Set the initial y-momentum equal to the final y-momentum. Divide Equation (2) by Equation (1) and solve for 0. Σpxi mcarV car = 3.75 x 104 kg. m/s Σpxf = (mcar + mvan) vf cos 0 = (4.00 × 10³ kg) vf cos ( ΣΡίγ (1) 3.75 x 104 kg tan 0 = Ꮎ +25.0 m/s = = (1.50 x 10³ kg)(25.0 m/s) mvan van = 5.00 x 104 kg. m/s = Σpfy = (mcar + mvan) vf sin 0 (4.00 x 10³ kg) vf sin 0 (2) 5.00 x 104 kg · m/s = (4.00 × 10³ kg) vf sin 0 53.1° x +20.0 m/s m/s = (4.00 × 10³ kg) vf cos 0 (2.50 x 10³ kg)(20.0 m/s) 5.00 x 104 kg. m/s 3.75 x 104 kg. m/s = 1.33
LEARN MORE
Vfx
REMARKS It's also possible to first find the x- and y-components and Vfy of the resultant velocity.
The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf
= √v 2
-1
Vfx² + Vfy and the inverse tangent function 0 tan-¹ (v/v). Setting up this alternate approach is a
simple matter of substituting vfx = vf cos and vfy = vf sin 0 in Equations (1) and (2).
QUESTION If the car and van had identical mass and speed, what would the resultant angle have
been?
I
magnitude
direction
PRACTICE IT
Use the worked example above to help you solve this problem. A car with mass 1.55 x 10³ kg traveling
east at a speed of 20.8 m/s collides at an intersection with a 2.55 x 10³ kg van traveling north at a speed
of 19.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after
the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together)
and assuming that friction between the vehicles and the road can be neglected.
m/s
° counterclockwise from the +x-axis
O
=
EXERCISE
A 3.22 kg object initially moving in the positive x-direction with a velocity of +4.99 m/s collides with and
sticks to a 2.23 kg object initially moving in the negative y-direction with a velocity of -2.56 m/s. Find the
final components of velocity of the composite object. (Indicate the direction with the sign of your answer.)
Vfx
Vfy
m/s
m/s
HINTS:
GETTING STARTED I
I'M STUCK!
Transcribed Image Text:LEARN MORE Vfx REMARKS It's also possible to first find the x- and y-components and Vfy of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, vf = √v 2 -1 Vfx² + Vfy and the inverse tangent function 0 tan-¹ (v/v). Setting up this alternate approach is a simple matter of substituting vfx = vf cos and vfy = vf sin 0 in Equations (1) and (2). QUESTION If the car and van had identical mass and speed, what would the resultant angle have been? I magnitude direction PRACTICE IT Use the worked example above to help you solve this problem. A car with mass 1.55 x 10³ kg traveling east at a speed of 20.8 m/s collides at an intersection with a 2.55 x 10³ kg van traveling north at a speed of 19.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. m/s ° counterclockwise from the +x-axis O = EXERCISE A 3.22 kg object initially moving in the positive x-direction with a velocity of +4.99 m/s collides with and sticks to a 2.23 kg object initially moving in the negative y-direction with a velocity of -2.56 m/s. Find the final components of velocity of the composite object. (Indicate the direction with the sign of your answer.) Vfx Vfy m/s m/s HINTS: GETTING STARTED I I'M STUCK!
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