wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After collision the 1 kg mass reverses its direction and moves with a speed of 2 ms1.
Which of the following statement(s) is/are correct for the system of these two masses?

A
Total momentum of the system after the collision is 3 Kg ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Momentum of 5 kg mass after the collision is 4 Kg ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Total kinetic energy of the system is 4 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Kinetic energy of the center of mass is 0.75 J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A Total momentum of the system after the collision is 3 Kg ms1
D Kinetic energy of the center of mass is 0.75 J



Given, m1=1 kg ; m2=5 kg

u1=u ; u2=0 ms1 and v1=2 ms1 ; v2=v

In collision, only internal forces are involved, hence momentum is conserved.

Pi=Pf

m1u1+m2u2=m1v1+m2v2

u=5v2 .......(1)

Since collision is elstatic, e=1

e=velocities of separationvelocities of approach=(v2v1)(u1u2)

1=v+2u

u=v+2 ........(2)

From (1) and (2) we get,

u=3 ms1 and v=1 ms1

Now, total moment of system is,

PT=m1v1+m2v2

=1×(2)+5×1=3 kg ms1

Now, total moment of 5 kg after the collision is,

P5 kg=m2v2=5 kg ms1

Total kinetic energy of the system,

KET=12m1v21+12m2v22=2+2.5=4.5 J

Kinetic energy of the center of mass is,

KEcm=12×msys×v2cm

=12×(1+5)×((1×u)+(5×0)1+5)2

=12×6×936=0.75 J

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) and (D) are the correct answer.
Why this question?
Key concept:
In an elastic collision total kinetic energy before and after collision is conserved.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon