A body of mass 22kg moving with velocity of 5m/s collides with a stationary body of mas 500g. if the two bodies move together after impact. calculate the magnitude of their common velocity
m1=22kg
u1=5m/s
m2=500g=0.5kg
u2=0m/s
v=v1+v2m/s
m1u1+m2u2=m1v+m2v
m1u1+m2u2=v(m1+m2)
22*5*0.5*0=v(22+0.5)
110=22+0.5v
v=110/22.5=4.888888
v=4.9m/s
v=4.9m/s
V=4.9m/s
To calculate the magnitude of their common velocity after the collision, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. Mathematically, this can be expressed as:
m1 * v1 + m2 * v2 = (m1 + m2) * v
where:
m1 is the mass of the first body (22 kg)
v1 is the initial velocity of the first body (5 m/s)
m2 is the mass of the second body (0.5 kg)
v2 is the initial velocity of the second body (0 m/s, as it is stationary)
v is the common velocity of both bodies after the collision.
Plugging in the given values into the equation:
(22 kg * 5 m/s) + (0.5 kg * 0 m/s) = (22 kg + 0.5 kg) * v
110 kg·m/s = 22.5 kg * v
Now we can solve for v by dividing both sides by 22.5 kg:
v = 110 kg·m/s / 22.5 kg
v ≈ 4.89 m/s
Therefore, the magnitude of their common velocity after the collision is approximately 4.89 m/s.