Question

At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/s)ˆx. One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the x-axis at x = 8.0 m. a) What is the mass of the particle at the origin? b) Calculate the total momentum of this system. c) What is the velocity of the particle at the origin?

Answer 1

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

Answer 2

(A) The mass of the particle at the origin is = 0.3kg

(B) The total momentum of this system is = 2 •i kgm/s

(C) The velocity of the particle at the origin is = 6.667 •i m/s

What is Momentum?

Given that,

At one instant,

The Center of mass is at 2m

$Xcm = 2m$

And velocity =5•i m/s

One of the particles is at the origin

M1=? X1 =0

The other has a mass$M2=0.1kg$

And it is at rest at position $X2= 8m$

A. Center of mass is given as per Question:

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

$2(M1+0.1) = 0.8$

$2M1 + 0.2 =0.8$

$2M1 = 0.8-0.2$

$2M1 = 0.6$

$M1 = 0.6/2$

$M1 = 0.3kg$

B. The Total momentum, this is an inelastic collision and it momentum after that collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

C. Although, The velocity of a particle at the origin

Then we are Using conversation of momentum:

When Momentum before the collision is equal to momentum after the collision

P(before) = M1 • V1 + M2 • V2

When we are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

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