Given
m1(mass of the first object): 55 Kg
m2 (mass of the second object): 55 Kg
v1 (velocity of the first object): 4.5 m/s
v2 (velocity of the second object): ?
m3(mass of the object dropped): 2.5 Kg
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
270 = 62.5 v2
v2 = 4.32 m/s
Answer:
Explanation:
Given
First bicycle travels 6.10 km due to east in 0.21 h
Suppose its position vector is 
After that it travels 11.30 km at 
east of north in 0.560 h
suppose its position vector is 
after that he finally travel 6.10 km due to east in 0.21 h
suppose its position vector is 
so position of final position is given by
t=0.21+0.56+0.21=0.98 h
For direction
w.r.t to x axis
Answer:
the answer has to be answer C
The heavier the object the faster it will fall. The lighter the object the slower to fall. You can test this for example with a rock and a piece of paper. The paper will take more time to for it to hit the ground rather than the rock which will be quicker.
