The momentum of the rolling ball will have less momentum than before the collision and the stationary ball will have more momentum after the collusion.
Answer:
Explanation:
we know angular velocity in terms of moment of inertia and angular speed
ω .... (1)
moment of inertia of rod rotating about its center of length b
........ .(2)
using v = ωr
where w is angular velocity
and r is radius of rod which is equal to b
so we get 2v = ωb
ω = 2v/b ................. (3)
here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v
put second and third equation in ist equation
×
so final answer will be 
Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
Answer:
Explanation:
⁶₃Li will have 3 protons and 3 neutrons .
mass of proton in amu = 1.00727 amu
mass of neutron in amu = 1.00866 amu
mass of lithium nucleus in amu = 6.01512 amu
mass defect = 3 ( 1.00727 + 1.00866 ) - 6.01512 amu
= .03267 amu
Binding energy = mass defect in amu x 931 Mev
= 30.41 MeV
binding energy per nucleon
no of nucleon = 3 + 3 = 6
binding energy per nucleon = 30.41 / 6 Mev
= 5.068 MeV .
Answer:
The total normal force acting on the system is approximately 58.8 N
Explanation:
The masses arranged in the stack are;
3 kg, 2 kg, and 1 kg
The mass of the stack system, m = 3 kg + 2 kg + 1 kg = 6 kg
Weight = The force of gravity on an object = m·g
Where;
m = The mass of the object
g = The acceleration due to gravity ≈ 9.8 m/s²
∴ The weight of the stack system, W ≈ 6 kg × 9.8 m/s² ≈ 58.8 N
The direction of the weight force = Perpendicular to the surface (acting downwards)
From Newton's third law of motion, the normal force acts perpendicular to the plane and it is equal in magnitude to the force acting perpendicular to the plane
∴ The magnitude of the total normal force acting on the system = The magnitude of the weight of the system ≈ 58.8 N
The (magnitude of the) total normal force acting on the system ≈ 58.8 N
