Block 1, of mass m1=1.50 kg, moves along a frictionless air track with speed v1=31.0m/s. It collides with block 2, of mass m2=11.0 kg, which was initially at rest. The blocks stick together after the collision.

A: Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.

B: Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.

C: what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

The type of collision in which the two blocks stick together after the

collision is a perfectly inelastic collision.

The correct responses are;

Reasons:

A: The magnitude of the total initial momentum, , is given as follows;

= m₁·v₁ + m₂·v₂

Where;

v₂ = The initial velocity of block 2 = 0 (block 2 is at rest)

Therefore;

= 1.50 × 31.0  + 11.0 × 0 = 46.5

The total initial momentum, = 46.5 kg·m/s

B: According to the principle of conservation of linear momentum, we have;

Total initial momentum, = Total final momentum,

The blocks stick together after the collision, therefore, the final velocity, , is the same

Which gives;

Total final momentum, = m₁· + m₂· = ·(m₁ + m₂)

Therefore, = , gives;

m₁·v₁ + m₂·v₂ = ·(m₁ + m₂)

46.5 = ·(1.50 + 11.0)

The final velocity, = 3.72 m/s

C: The change in kinetic energy, ΔK = is given as follows;

The kinetic energy, K.E. =

Therefore;

ΔK = = 86.49 - 720.25 = -633.76

The change in kinetic energy, ΔK = -633.76 kJ

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A) 46.5 kg m/s

B) 3.72 m/s

C) -634.3 J

Explanation:

A)

The momentum of an object is a vector quantity given by:

where

m is the mass of the object

v is its velocity

Momentum is a vector quantity, so if we want to calculate the total momentum of a system of two blocks, we have to add the individual momenta vectorially.

Here we have:

- Momentum of block 1:

where

is the mass

is the velocity

- Momentum of block 2:

where

is the mass of block 2

is its velocity

So, the total momentum at the beginning is

B)

We can solve this problem by using the law of conservation of momentum.

In fact, for an isolated system (=no external forces), the total momentum of the system must be conserved.

Therefore, here we can write that the final total momentum is equal to the initial momentum calculated in part a):

And the final momentum can be written as:

where

is the mass of block 1

is the mass of block 2

is the final velocity of the two blocks sticking together

Combining the two equations and solving for vf, we find:

C)

Here we want to calculate the change in kinetic energy of the system.

The kinetic energy of an object is given by

where

m is the mass of the object

v is its speed

Before the collision, the total kinetic energy of the system is just the kinetic energy of block 1 (which is the only one in motion), so:

While the total kinetic energy after the collision is:

Therefore, the change in kinetic energy is

So, the system has lost 634.3 J of energy.