Answer:
D) momentum of cannon + momentum of projectile= 0
Explanation:
The law of conservation of momentum states that the total momentum of an isolated system is constant.
In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:
But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:
So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:
Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle,
Coefficient of kinetic friction,
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:
Now, the speed is given by the third eqn of motion with initial velocity being zero:
where
u = initial velocity = 0
Thus