Two hockey players have a total momentum of +200 kg x m/s before a collision (+ is to the right). After the collision, they move
1 answer:
Total momentum after the collision: +200 kg m/s to the right
Explanation:
We can answer this question by using the law of conservation of momentum, which states that for an isolated system (=no external forces acting on the system), the total momentum is conserved.
Mathematically,
where
is the total momentum before the collision
is the total momentum after the collision
In this problem, the system consists of two hockey players. Before the collision, their total momentum is
(to the right)
Therefore, according to the law of conservation of momentum, their total momentum after the collision must be the same:
And given that the sign is +, the direction is still the same, therefore to the right.
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Answer:
1034.78 m
Explanation:
The shortest distance is the displacement of the car from initial position to final position.
Displacement of body is given using Newton's equations of motion.
Given:
Initial velocity, m/s
Final velocity, m/s( As the car stops in its final position)
Coefficient of static friction,
Acceleration due to gravity,
Now, when brakes are applied, only friction acts on the body in a direction opposite to that of its motion.
The acceleration of the car when friction acts the stopping force is given as:
.
The acceleration is negative as it reduces the velocity of the motion and acts in the direction opposite to that of the motion.
Plug in 0.71 for and 9.8 m/s² for
. Solve for
.
So, .
Now, displacement of the car is given using the following equation of motion:
Here, is the displacement of the racing car.
Plug in 120 m/s for , 0 m/s for
, -6.958 m/s² for
. Solve for
. This gives,
Therefore, the shortest distance in which she can brake and stop is 1034.78 m.