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
Answer: Got It!
<em>Explanation:</em> Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
Answer:
kg m/s
Explanation:
e = Charge = C
V = Voltage = 
c = Speed of light = m/s
Momentum is given by
The unit of MeV/c in SI fundamental units is kg m/s
