Physics, asked by sakshambagri81patljr, 1 year ago

An object of mass 1.5kg travelling in a straight line with a velocity of 5m/s collides with a wooden block of mass 5kg resting on the floor. This object sticks with the wooden block after collision and both move together in a straight line. Calculate:
(a)the total momentum before collision.
(b) total momentum after collision
(c) the velocity of the combination of these objects after collision

Answers

Answered by aqibkincsem
57

"Mass of the moving object, m1 = 1.5 kg and its initial velocity, u1 = 5 m/s.

Mass of the wooden block, m2 = 5 kg, and its initial velocity, u2 = 0 m/s.

We need to find total momentum before and after the collision of both objects. Therefore m1u1 + m2u2 = m1v1 + m2v2 => v = 1.15 m/s

Before collision, total momentum 7.5 kgm/s and after collision 7.47 kgm/s."

Answered by MahathiR1
4

Answer:

2.25 m/s

Explanation:

Here, m  

1

=10g=  

1000

10

=0.01kg

m  

2

=20g=0.02kg

u  

1

=3ms  

−1

;u  

2

=2ms  

−1

 

v  

1

=2.5ms  

−1

;v  

2

=?

Total momentum of both the spheres before collision =m  

1

u  

1

+m  

2

u  

2

 

=0.01×3+0.02×2=0.07kgms  

−1

 

Total momentum of both the spheres after collision

=m  

1

v  

1

+m  

2

v  

2

 

=0.01×2.5+0.02v  

2

=0.025+0.02v  

2

 

Now, according to the law of conservation of momentum,

Total momentum after collision =Total momentum before collision

∴0.025+0.02v  

2

=0.07

or 0.02v  

2

=0.07−0.025=0.045

or v  

2

=  

0.02

0.045

=2.25ms  

−1

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