Two blocks have mass m and m=2.77 m, respectively. a light spring is attached to one of them, and the blocksare pushed together, compressing the spring between them. they are secured, in this compressed state, by a cord. they are then placed, at rest, on a horizontal frictionless surface. the cord holding them together is burned, after which the block of mass m moves to the right with a speed of 1.38 m/s. what is the speed of the block of mass m? answer in units of m/s

3.8226 m/s

Explanation:

Momentum is conserved hence

equating momentum, p

p (before) = p (after)

0 = (m1v1)+(M2v2)  but m2=2.77m where m and v are the masses and velocity respectively

0 = (mv1) +(2.77m)(1.38)

(mv1)=-(2.77m)(1.38)

(v1)=-(2.77)(1.38)

Solving v1

v1 = - 3.8226 m/s

= 3.8226 m/s, to the left

According to the law of conservation of linear momentum, the linear

momentum of the blocks is conserved (remains the same).

Reasons:

Mass of the two blocks are; m and M = 2.77·m

Location of the spring = Between the blocks

Speed of the block of mass M after the cord is burned, v₂ = 1.38 m/s

Required:

The speed of the block of mass m

Solution:

According to the law of conservation of linear momentum, we have;

Initial momentum of the blocks = Final momentum after the cord is burned

The initial velocity of the blocks when secured, v₀ = 0

v₁ = The velocity of the block of mass m

Therefore;

(m + M) × 0 = m·v₁ + M × 1.38 = m·v₁ + 2.77·m × 1.38 = m·v₁ + 3.8226·m

(m + M) × 0 = 0

Which gives;

(m + M) × 0 = 0 = m·v₁ + 3.8226·m

m·v₁ + 3.8226·m = 0

m·v₁ = -3.8226·m

v₁·m = -3.8226·m

By comparison of the coefficient of m in the above equation, the velocity of the block of mass m, v₁ = -3.8226 m/s.

Therefore, the magnitude of the velocity of block m, |v₁| = 3.8226 m/s

The direction in which the block of mass m moves = opposite to the direction of the block of mass M.

Given that speed is a scalar quantity, we have;

Learn more about the law of conservation of linear momentum here:

link

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

v=ir = 3x15 =45 v