Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 2, Problem 2.1P

Two blocks of mass 0.1 kg and 0.2 kg approch each other on a frictionless surface at velocities of 0.4 and 1 m/s, respectly. If the blocks collide and remain together, calculate their joint after the collision.

Expert Solution & Answer
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To determine

The joint velocity after the collision Two blocks of mass 0.1 kg and 0.2 kg approach each other on a frictionless surface at velocities of 0.4 and 1 m/s, respectively, if the blocks collide and remain together?

Explanation of Solution

Given info:

  m1=0.1kgm2=0.2kgv1=0.4m/sv2=1m/s

Formula used:

By law of conservation of momentum,

  m1v1+m2v2=MV

Calculation:

Here momentum is conserved

Assume mass of first block and second block as m1 and m2 , and their respective velocities as v1 and v2 .After collision let the velocity and mass be M and V.

We have v1 and v2 are in opposite direction, so

  m1v1m2v2=MV0.1×0.40.2×1=0.3×VV=0.040.20.3=0.160.3=0.53m/s

Conclusion:

Thus, the velocity of blocks after collision is 0.53 m/s towards left.

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