Two blocks collide on a frictionless surface, as shown above. They have a combined mass of 10 kg and a speed of 2.5 m/s. Before the collision, one of the blocks was at rest. This block had a mass of 8 kg. What was the speed of the second block?

12.5 m/s

Explanation:

Excuse my scribbles!

I had to work backwards using the inelastic collision formula for this problem.

Step 1: Substitute in the values provided in the problem

2.5=(8*0)+(?*?)/(8+?)

Step 2: Subtract block 1's mass from the combined mass to determine block 2's mass

10-8=2     Block 2's mass is 2.

2.5=(8*0)+(2*x)/(8+2)    

Now simplify.

2.5=(2*x)/(8+2)

2.5=2x/10

Step 3: Multiply both sides by the reciprocal

(5)2.5=2x/10(5)

12.5=x

Answer is checked in the attached images!

answer; equivalent to a one(1) horsepower motor;

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s