A 306 g cart moves on a horizontal, frictionless surface with a constant speed of 14.2 cm/s. A 76.3 g piece of modeling clay is
1 answer:
To solve this problem it is necessary to apply the concepts related to the conservation of the momentum for an inelastic shock. Mathematically this can be described as
Where,
m_{1,2} = Mass of each object
v_{1,2} = Initial velocity of each object
V_f = Final Velocity
If we assign the value of mass 1 and speed 1 to the cart and the variable of mass 2 to clay (which is at rest) we will have:
Therefore the final speed of the system would be 11.36cm/s
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Answer:
A) mass = 3121.58 kg
B) tension = 25940.37 N
C) tension = 25940.37 N (tension on both sides will be the same)
Explanation:
Weight of elevator = 22500 N
Distance = 6.75 m
Time = 3 sec
Since it started from rest, initial speed is zero.
Using Newton's equation of motion we have,
S = ut + 0.5at^2
S = distance covered = 6.75 m
t = time = 3 s
a = acceleration upwards
u = initial velocity = 0
Substituting values, we have,
6.75 = 0(3) + (0.5 x a x 3^2)
6.75 = 4.5a
a = 6.75/4.5 = 1.5 m/s^2 (acceleration of the elevator upwards)
Mass of the elevator = Weight/g
Where g = acceleration due to gravity 9.81 m/s
Mass = 22500/9.81 = 2293.58 kg
From the image below we solve from
T - 22500 = ma
T - 22500 = 2293.58 x 1.5
T - 22500 = 3440.37
T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)
On the other side,
mg - T = ma
9.81m - 25940.37 = 1.5m
(9.81 - 1.5)m = 25940.37
8.31m = 25940.37
m = 3121.58 kg (mass of counter weight)
See image below
