Answer:
Magnitude of impulse, |J| = 4 kg-m/s
Explanation:
It is given that,
Mass of cart 1, 
Mass of cart 2, 
Initial speed of cart 1, 
Initial speed of cart 2, 
(stationary)
The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.
V = 1 m/s
The magnitude of the impulse exerted by one cart on the other is given by:
J = -4 kg-m/s
or
|J| = 4 kg-m/s
So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.
Answer:
0.22 m
Explanation:
We are told that the driver can survive an acceleration of 50g only if the collision lasts no longer than 30 ms. So,
The acceleration is
where the negative sign is due to the fact that this is a deceleration, since the driver comes to a stop in the collision.
First of all, we can find what the initial velocity of the car should be in this conditions by using the equation:
And since the final velocity is zero, v=0, and solving for u,
And now we can find the corresponding distance travelled using the equation:
Answer:
(A) 374.4 J
(B) -332.8 J
(C) 0 J
(D) 41.6 J
(E) 351.8 J
Explanation:
weight of carton (w) = 128 N
angle of inclination (θ) = 30 degrees
force (f) = 72 N
distance (s) = 5.2 m
(A) calculate the work done by the rope
- work done = force x distance x cos θ
- since the rope is parallel to the ramp the angle between the rope and
the ramp θ will be 0
work done = 72 x 5.2 x cos 0
work done by the rope = 374.4 J
(B) calculate the work done by gravity
- the work done by gravity = weight of carton x distance x cos θ
- The weight of the carton = force exerted by the mass of the carton = m x g
- the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.
work done by gravity = 128 x 5.2 x cos 120
work done by gravity = -332.8 J
(C) find the work done by the normal force acting on the ramp
- work done by the normal force = force x distance x cos θ
- the angle between the normal force and the ramp is 90 degrees
work done by the normal force = Fn x distance x cos θ
work done by the normal force = Fn x 5.2 x cos 90
work done by the normal force = Fn x 5.2 x 0
work done by the normal force = 0 J
(D) what is the net work done ?
- The net work done is the addition of the work done by the rope, gravitational force and the normal force
net work done = 374.4 - 332.8 + 0 = 41.6 J
(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal
- work done by the rope= force x distance x cos θ
- the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp
work done = 72 x 5.2 x cos 20
work done = 351.8 J
Answer:
26.67 m/s
Explanation:
From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.
p=mv where p is momentum, m is the mass of object and v is the speed of the object
Initial momentum
The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero
Final momentum
Answer:
Obviously Lengthen... 
or 
Explanation:
As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...
