vector A has magnitude 12 m and direction +y
so we can say
vector B has magnitude 33 m and direction - x
Now the resultant of vector A and B is given as
now for direction of the two vectors resultant will be given as
so it is inclined at 160 degree counterclockwise from + x axis
magnitude of A and B will be
so magnitude will be 35.11 m
The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and m₂ will be u₁=5 and u₂=0 m/s respectively,
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
m₁u₁+m₂u₂=(m₁+m₂)v
M×5+3M×0=[M+3M]v
The final velocity is found as;
V=51.25 m/s
The velocity of block A is found as;
Hence, the final velocity of the block A will be 2.5 m/sec.
To learn more about the law of conservation of momentum, refer;
#SPJ4
Answer:
Explanation:
The rotational kinetic energy when the cylinder is with the rope is:
where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:
(1)
For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:
Finally, by replacing in (1):
hope this helps!!