consider the right direction as positive and left direction as negative.
M = mass of the ball = 5 kg
m = mass of stone = 1.50 kg
= initial velocity of the ball before collision = 0 m/s
= initial velocity of the stone before collision = 12 m/s
= final velocity of the ball after collision = ?
= final velocity of the stone after collision = - 8.50 m/s
using conservation of momentum
M + m = M + m
(5) (0) + (1.5) (12) = 5 + (1.50) (- 8.50)
= 6.15 m/s
h = height gained by the ball
using conservation of energy
Potential energy gained by ball at Top = kinetic energy at the bottom
Mgh = (0.5) M
(9.8) h = (0.5) (6.15)²
h = 1.93 m
Answer:
After throwing the object the, the velocity of the man is 13.98 m/s
Explanation:
Given:
Let,
mass of man, m1 = 74 kg
mass of box, m2 = 13 kg
Initial velocity u = 11 m/s (initially both are together hence initial velocity will be same for both)
Final velocity of man = v1
Final velocity of box = v2 = -6 m/s (the velocity is recoil velocity therefore it is negative)
To Find:
Final velocity of man,after throwing the object = v1 = ?
Solution:
Recoil velocity:
It is the backward velocity experienced.
Here recoil velocity is the backward velocity experience while throwing the box behind.Hence the velocity of the box is negative 6 m/s.
The recoil velocity is the result of conservation of linear momentum of the system. Therefore we will follow the law of conservation of momentum.
Law of conservation of momentum :
Total momentum of an isolated system before collision is always equal to total momentum after collision
substituting the values which are given above we get
Therefore, After throwing the object the, the velocity of the man is 13.98 m/s
Answer:
B. 8.0m/s
Explanation:
Distance between waves = 50m
T = 6.25secs
Speed = 50m/6.25secs
= 8.0m/s
