Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.
Explanation:
The momentum of a car (an object) is
p= mv
where
m is =the mass of the object( in this case car)
v is its= velocity
While the kinetic energy is is given by the formulae
K=1/2mv²
To determine how momentum and kinetic energy of the car changes when the speed of the object triples, We have that the new velocity,
v¹= 3v
So that the momentum change becomes
p¹=mv¹=m (3v)= 3mv
mv=p
therefore p¹= 3p
we can see that the momentum also triples.
And the kinetic energy change becomes
K¹=1/2m(v¹)²= 1/2m (3v)²
= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²
1/2mv²=K
K¹= Kinetic energy = 9k
but Kinetic energy increases 9 times
1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the 
and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:
Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:
It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.
The answer would be solubility
True if you look up the question Is velocity speed in a certain direction you would’ve gotten the answer but I’m pretty sure it’s true
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
