Help me plz
.show working

1. 5m/s

Explanation:

1. We can apply the law of conservation of momentum here;

m1*u1 = m1*v1+ m2*v2

m1*u1 is 5 since the balls have equal masses-- m1*u1=5=m1(v1 + v2)

similarly the balls have a velocity of 5m/s each since v1+v2 = 10m/s

(answer choice c)

your can similarly calculate the second problem

For both questions, you have to remember:

-- momentum of one object = (mass) x (velocity)

-- (total momentum after the collision) = (total momentum before it)

and

-- keep the directions of the velocities straight.

#1:

What's the total momentum before the collision ?

Ball-A ... (0.5 kg) x (10 m/s forward) = 5 kg-m/s forward

Ball-B ... (0.5 kg) x (zero)  =  zero

Sum before the collision = 5 kg-m/s forward

The question says that both balls move away from the collision together.

So their velocities are the ame speed in the same direction.

What's the total momentum after the collision ?

Ball-A ... (0.5 kg) x (v m/s)

Ball-B ... (0,5 kg) x (v m/s)The trick to this one is to keep straight on the direction of 'velocity'.

But it has to be equal to the sum before the collision, so

(1 kg) x (v m/s) = 5 kg-m/s forward

v = 5 m/s forward (choice - C)

#2:

What's the total momentum before the collision ?

Ball-A ... (6 kg) x (10 m/s forward) = 60 kg-m/s forward

Ball-B ... (2 kg) x (5 m/s backward) = 10 kg-m/s backward

Sum before the collision = 50 kg-m/s forward

The question says that both balls stick together after the collision.

So now we have one object of 8 kg, moving with a velocity of v .

Total momentum after the collision:

One object ... (8 kg) x (v m/s)

But it has to be equal to the sum before the collision, so

(8 kg) x (v m/s) = 50 kg-m/s forward

v = 6.25 m/s forward (choice - B)

not moving. pos accel,

this could be newton#'s law of cooling. the cooling curve is exponential