Answer:
0.167m/s
Explanation:
According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.
Given momentum = Maas × velocity.
Momentum of glider A = 1kg×1m/s
Momentum of glider = 1kgm/s
Momentum of glider B = 5kg × 0m/s
The initial velocity of glider B is zero since it is at rest.
Momentum of glider B = 0kgm/s
Momentum of the bodies after collision = (mA+mB)v where;
mA and mB are the masses of the gliders
v is their common velocity after collision.
Momentum = (1+5)v
Momentum after collision = 6v
According to the law of conservation of momentum;
1kgm/s + 0kgm/s = 6v
1 =6v
V =1/6m/s
Their speed after collision will be 0.167m/s
Answer:
how ever many layers it has i think
Explanation:
1.velocity and acceleration
2.
3.inertia
4.
5.speed
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
PITCH
please mark brainliest!!!
