In elastic
collision, both the kinetic energy and momentum are conserved. Conservation
means that both the kinetic energy and momentum will have the same values
before and after elastic collision.
<span>As the
object A has low mass than object B. Hence upon collision, object B moves
forward, while object A will move backward. So option "C" is correct. </span>
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:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

Answer:
heat required in pan B is more than pan A
Explanation:
Heat required to raise the temperature of the substance is given by the formula

now we know that both pan contains same volume of water while the mass of pan is different
So here heat required to raise the temperature of water in Pan A is given as


Now similarly for other pan we have


So here by comparing the two equations we can say that heat required in pan B is more than pan A
Answer:
a = the lowest critical speed of the shaft 882.81 rad/s
b = new diameter 0.05m or 50mm
c = critical speed 1765.62rad/s
Explanation:
see the attached file