answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Olin [163]
2 years ago
8

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

and hit a stationary green bumper car of mass 150 kg. The red, blue, and green bumper cars all combine.
What is the final velocity of the combined bumper cars?
A .93m/s
B 1.2 m/s
C 1.7m/s
D 2.7m/s
Please Help!!!
Physics
2 answers:
Illusion [34]2 years ago
6 0

Answer: B. 1.2 m/s

Explanation:

First we would find the velocity with which red car and blue car move when they combine after first collision.

Using law of conservation of momentum,

Initial momentum = final momentum

Mass of red Car, Mr = 225 kg

Mass of blue car, Mb = 180 kg

Mass of green car, Mg = 150 kg

Initial velocity of red car, Ur = 3.0 m/s

Initial velocity of blue car, Ub = 0

Initial velocity of green car, Ug = 0

collision 1:

Mr Ur + Mb Ub = (Mr+Mb) V

where, V is the final velocity with which red and blue car moves.

225 × 3 + 0 = (225 + 180 ) V

⇒ V = 675/ 405 = 1.67 kg.m/s

collision 2:

These cars  collide with the green car. Let the final velocity with which the three cars move together after collision: V'

(Mr+Mb) V + Mg Ug = (Mr+Mb+Mg) V'

675 + 0 = 555 V'

⇒V' = 1.2 m/s

Thus, the correct option is B.

Alex Ar [27]2 years ago
4 0

Given


m1(mass of red bumper): 225 Kg


m2 (mass of blue bumper): 180 Kg


m3(mass of green bumper):150 Kg


v1 (velocity of red bumper): 3.0 m/s


v2 (final velocity of the combined bumpers): ?




The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:


Pa= Pb


Where Pa is the momentum before collision and Pb is the momentum after collision.


Now applying this law for the above problem we get


Momentum before collision= momentum after collision.


Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s


Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

Now we know that Momentum before collision= momentum after collision.


Hence we get


1215 = 555 v2


v2 = 2.188 m/s


Hence the velocity of the combined bumper cars is 2.188 m/s

You might be interested in
A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pres
marishachu [46]

Answer:

C) The pressure reading stays the same.

Explanation:

3 0
2 years ago
Read 2 more answers
One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the
sergeinik [125]
<span>Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20</span>
4 0
3 years ago
Read 2 more answers
The van of Hans and Frans is stuck on slippery ice, so they must get out and move it by hand. Hans pushes
Irina18 [472]

Answer: a= ff+fh/m

Explanation: bc khan academy said it was d. a=ff +fh/m

4 0
2 years ago
Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea’s reference frame is
BartSMP [9]

Answer:  Sean is standing still, and Rhea is running toward Sean while   kicking the ball

Explanation: Your welcome :)

5 0
2 years ago
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45
Helga [31]

Answer:

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

8 0
2 years ago
Other questions:
  • You do 174 J of work while pulling your sister back on a swing, whose chain is 5.10 m long, until the swing makes an angle of 32
    8·1 answer
  • An air-filled 20-μf capacitor has a charge of 60 μc on its plates. how much energy is stored in this capacitor?
    8·1 answer
  • An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
    7·1 answer
  • As she was trying to study, Tanisha asked her roommate to lower the radio. Her roommate had turned the radio up originally from
    7·1 answer
  • A floating ice block is pushed through a displacement d = (14 m) i hat - (11 m) j along a straight embankment by rushing water,
    15·1 answer
  • A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
    10·1 answer
  • A hydroelectric dam holds back a lake of surface area 3.0×106m2 that has vertical sides below the water level. The water level i
    6·1 answer
  • g A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to th
    5·1 answer
  • A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the r
    7·1 answer
  • A dragster crosses the finish line with a velocity of 140m/s . Assuming the vehicle maintained a constant acceleration from star
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!