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irga5000 [103]
3 years ago
9

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 77.0 kg, is moving to the right at 4.00

m/s, while the other, of mass 66.0 kg, is moving to the left at 2.50 m/s.
What is the magnitude of the velocity of these skaters just after they collide?
What is the direction of the velocity of these skaters just after they collide?
a. to the left
b. to the right
Physics
1 answer:
Alex777 [14]3 years ago
5 0

Answer:3.31m/s², to the right

Explanation:

According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.

Momentum = mass × velocity

M1 and M2 be the masses of the first and second skaters respectively

Let u1 and u2 be the velocities of the first and second skaters respectively.

v be their common velocity after collision

M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²

According to the law we have

M1u1 + M2u2 = (M1+M2)v

77(4) + 66(2.5) = (77+66)v

308 + 165 = 143v

V = 473/143

V = 3.31m/s²

Their velocity after collision will become 3.31m/s²

They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.

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6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant
Sindrei [870]

Answer:

Explanation:

a )

momentum of baseball before collision

mass x velocity

= .145 x 30.5

= 4.4225 kg m /s

momentum of brick after collision

= 5.75 x 1.1

= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball  before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

 kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J

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strojnjashka [21]
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1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15
kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity=\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 m/s^2

Distance covered by the car,s = 75 m

Initial velocity of the car ,v_i = 6 m/s

Final velocity of thre car ,v_f=?

Using third equation of motion:

v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2

v_{f}=14.69 m/s

The final velocity of the car at the end of 75 m is 14.69 m/s

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You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of
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