Two blocks of masses M and 3 M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. a. What is the speed of the block of mass M?
b. Find the original elastic energy in the spring if M = 0.30 kg.

-6 m/s

7.2 J

Explanation:

Given that

Initial velocity, u of both bodies = 0

Initial momentum = 0

Final velocity, v of the 3M block = 2 m/s

Mass M of the block A = M

Mass M' of the block B = 3M

Using the law of conservation of momentum, we know that

M * V(a) + 3M * V(b) = 0

Remember that the velocity of the block 3M block is V(b) = 2 m/s, so then

M.V(a) = - 3M * 2

V(a) = -6M/M

V(a) = -6 m/s

b

If mass M = 0.3 kg, then

K.E = 1/2 * M * V(a)² + 1/2 * M' * V(b)²

If we substitute values, we have

K.E = 1/2 * 0.3 * (-6)² + 1/2 * (3 * 0.3) * 2²

K.E = 0.15 * 36 + 2 * 0.9

K.E = 5.4 + 1.8

K.E = 7.2 J

I = 0.25 [amp]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.

V = I*R

where:

V = voltage [Volt]

I = amperage or current [amp]

R = resistance [ohm]

Since all resistors are connected in series, the total resistance will be equal to the arithmetic sum of all resistors.

Rt = 2 + 8 + 14

Rt = 24 [ohm]

Now clearing I for amperage

I = V/Rt

I = 6/24

I = 0.25 [amp]