Block B (mass 7.50 kg ) is at rest at the edge of a smooth platform, 2.60 m above the floor. Block A (mass 4.00 kg ) is sliding with a speed of 8.00 m/s along the platform toward block B. A strikes B and rebounds with a speed of 2.00 m/s. The collision projects B horizontally off the platform.

What is the speed of B just before it strikes the floor?

speed of block B after collision is 5.33 m/s

speed of block B before strike speed 8.9 m/s

Explanation:

given data

mass = 7.50 kg

height = 2.60 m

speed =8.00 m/s

rebound speed = 2.00 m/s.

solution

we will apply here conservation of momentum that is

m1v1  + m2v2  = m1vr + m2v    .................1

4 × 8 + 7.50 × 0 = 4 × (-2) + 7.50 × v  

solve it we get

v = 5.33 m/s

so here speed of block B after collision is 5.33 m/s

and

and now by conservation of energy

we get here strike speed that is

PE(i) + KE(i) = PE(f) + KE(f)   ..............2

m×g×h + 0.5 × m × (v(i))² = 0 + 0.5 ×m×(v(f))²  

m ( g×h + 0.5×(v(i))²) = 0.5 ×m×(v(f))²  

9.8 × 2.60 + 0.5 × 5.33² = 0.5  × (v(f))²  

solve it we get

v(f)² =  

v(f) =

v(f) = 8.9 m/s

so speed of block B before strike speed 8.9 m/s

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

the right answer is 0.3 meters per second