Need help due in 30 mins​

1. v₂ = 12.43 m/s

2. I₁ = - 80 N.s

3. I₂ = 80 N.s

Explanation:

Consider the following data given in the question:

m₁ = mass of the Cart 1 = 16 kg

m₂ = mass of the Cart 2 = 7 kg

u₁ = velocity of cart 1 before collision = 4 m/s

u₂ = velocity of cart 2 before collision = 1 m/s

v₁ = velocity of cart 1 after collision = - 1 m/s

v₂ = velocity of cart 2 after collision = ?

1.

Applying the law of conservation of momentum:

v₂ = 12.43 m/s

2.

Impulse of cart 1 = I₁ = m₁(v₁ - u₁)

I₁ = (16 kg)(- 1 m/s - 4 m/s)

I₁ = - 80 N.s

3.

Impulse of cart 2 = I₂ = m₂(v₂ - u₂)

I₂ = (7 kg)(12.43 m/s - 1 m/s)

I₂ = 80 N.s

i will answer your question if you answer mine

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft