Answer:
Part a)
Part b)
Part C)
Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as
now we will have
Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have
Part C)
By momentum conservation we will have
Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
The answer should be B. According to the conservation of energy, the energy cannot be created nor destroyed, but it can be transformed. Since the object is moving down, that means its height is decreasing, causing the potential energy decreasing and the kinetic energy increasing to fulfill the conservation law.
Answer:
A) greater
Explanation:
acceleration is calculated by dividing velocity over time..so by calculating, you find acceleration of A is greater than that of B
Answer:
0.167m/s
Explanation:
According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.
Given momentum = Maas × velocity.
Momentum of glider A = 1kg×1m/s
Momentum of glider = 1kgm/s
Momentum of glider B = 5kg × 0m/s
The initial velocity of glider B is zero since it is at rest.
Momentum of glider B = 0kgm/s
Momentum of the bodies after collision = (mA+mB)v where;
mA and mB are the masses of the gliders
v is their common velocity after collision.
Momentum = (1+5)v
Momentum after collision = 6v
According to the law of conservation of momentum;
1kgm/s + 0kgm/s = 6v
1 =6v
V =1/6m/s
Their speed after collision will be 0.167m/s
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
