13. A block with a mass m1 = 2.0-kg moves with a speed of v1 = 8.0 m/s on a fric

Question

13. A block with a mass m1 = 2.0-kg moves with a speed of v1 = 8.0 m/s on a frictionless table approaching a second block with a mass m2 = 6.0-kg and a speed of 2.0 m/s.

(i) If this is a collision where after the event the two blocks are deformed into one large mass, then what is the
velocity (speed and direction) of that mass?

(ii) If instead this is a collision where the two blocks collide and conserve kinetic energy in the process, each
moving away at its own velocity, then what is the final velocity of each of the blocks?

A block with a mass m1 = 2.0-kg moves with a speed of v1 = 8.0 m/s on a frictionless table approaching a second block with a mass m2 = 6.0-kg and a speed of 2.0 m/s. If this is a collision where after the event the two blocks are deformed into one large mass, then what is the velocity (speed and direction) of that mass? If instead this is a collision where the two blocks collide and conserve kinetic energy in the process, each moving away at its own velocity, then what is the final velocity of each of the blocks?

Explanation / Answer

1. conservation of momentum,

2*8 - 2*6 = (2+6)*v

4=8v

v=0.5 m/s

2.applying conservation of momentum,

2*8 - 6*2 = 2*x+ 6*y

4= 2x+6y

x+3y=2

squaring both the sides

x^2+9y^2+6xy=4 -------------- (a)

also

applying conservation of kinetic energy

1*64+3*4=1*x^2+3y^2=76 ----------------- (b)

solving above two equations we have

76-3y^2+9y^2+6y(2-3y)-4=0

6y^2+12y-18y^2+72=0

9y^2-12y-72=0

y=3.57 or -2.24

y=3.57m/s

x=8.71m/s

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