Two figure skaters are coasting in the same direction, with the leading skater moving at 4.7 m/s and the trailing skater moving at 5.7 m/s. When the trailing skater catches up with the leading skater, he picks her up without applying any horizontal forces on his skates. If the trailing skater is 40% heavier than the 50 kg leading skater, what is their speed (in m/s) after he picks her up?
conserve momentum. Note that the actual value of m does not matter, since we have the relative masses.
1.4m*5.7 + m*4.7 = 2.4m*v
7.98+4.7 = 2.4v
v = 5.28 m/s
THANK YOU!!!
Well, it seems like these skaters are in quite the "coasting" situation! Although the trailing skater might have caught up to the leading skater, it's time for some physics fun!
We know the speed of the leading skater is 4.7 m/s, and the speed of the trailing skater is 5.7 m/s. Now, because the trailing skater picks up the leading skater, we can assume that the momentum is conserved in the horizontal direction.
Let's consider the momentum of the two skaters before the pick-up:
Momentum of the leading skater = mass x velocity
Momentum of the trailing skater = mass x velocity
Since their momentum is conserved even after the pick-up, we can write:
Momentum before = Momentum after
The sum of their momenta before the pick-up is:
(50 kg) x (4.7 m/s) + (1.4 x 50 kg) x (5.7 m/s) = 50 kg x v1 + (1 + 0.4) x 50 kg x v2
Solving this equation, we find that the final speed (v2) after the pick-up is approximately 4.64 m/s.
So after the trailing skater picks up the leading skater, they'll be gliding along with a speed of 4.64 m/s. They might even have time for a quick spin on the ice before they bust out some sweet dance moves!
Let's assume the leading skater's speed (v1) is 4.7 m/s, the trailing skater's speed (v2) is 5.7 m/s, and the initial mass of the leading skater (m1) is 50 kg.
Since the trailing skater catches up with the leading skater without applying any horizontal forces, the net force acting on both skaters must be zero. This means the total momentum before picking her up is equal to the total momentum after picking her up.
The momentum of an object is calculated by multiplying the mass of the object by its velocity.
Initial momentum of leading skater (p1) = m1 * v1
Final momentum of leading skater (p1') = (m1 + 0.4m1) * v' (where v' is the combined velocity after picking her up)
The initial momentum of the trailing skater (p2) = m2 * v2 (where m2 is the mass of the trailing skater)
Final momentum of the trailing skater (p2') = (m2 + 0.4m1) * v' (where v' is the combined velocity after picking her up)
According to the law of conservation of momentum, the total momentum before picking her up is equal to the total momentum after picking her up:
p1 + p2 = p1' + p2'
Substituting the momentum equations, we have:
m1 * v1 + m2 * v2 = (m1 + 0.4m1) * v' + (m2 + 0.4m1) * v'
Simplifying the equation:
50 kg * 4.7 m/s + m2 * 5.7 m/s = (50 kg + 0.4 * 50 kg) * v' + (m2 + 0.4 * 50 kg) * v'
Since the trailing skater is 40% heavier than the leading skater, m2 = 1.4 * m1
50 kg * 4.7 m/s + 1.4 * 50 kg * 5.7 m/s = (50 kg + 0.4 * 50 kg) * v' + (1.4 * 50 kg + 0.4 * 50 kg) * v'
Simplifying further:
235 kg * m/s + 399 kg * m/s = 90 kg * v' + 140 kg * v'
634 kg * m/s = 230 kg * v' + 140 kg * v'
Dividing both sides of the equation by 10:
63.4 m/s = 23 kg * v' + 14 kg * v'
Now we can solve for the combined velocity after picking her up (v').
63.4 m/s - 14 kg * v' = 23 kg * v'
Subtracting 23 kg * v' from both sides:
40.4 m/s = 9 kg * v'
Dividing both sides by 9 kg:
v' = 40.4 m/s / 9 kg
Therefore, their speed after he picks her up is approximately 4.49 m/s.
To solve this problem, we need to apply the principle of conservation of linear momentum. The total momentum of the system before the trailing skater picks up the leading skater should be equal to the total momentum of the system after they are combined.
Let's denote the mass of the leading skater as m and the mass of the trailing skater as 1.4m (since the trailing skater is 40% heavier than the leading skater). Let's also denote the initial velocities of the leading and trailing skaters as v1 and v2, respectively. After the trailing skater picks up the leading skater, they move together as a single combined mass.
The total initial momentum of the system is given by:
m1v1 + m2v2
The total final momentum of the system is given by:
(m1 + m2)V
where V is the common velocity of the combined mass after they are picked up.
Since we know the initial velocities (v1 = 4.7 m/s and v2 = 5.7 m/s) and the mass of the leading skater (m = 50 kg), we can write the equation as:
m(4.7) + 1.4m(5.7) = (m + 1.4m)V
Simplifying the equation:
4.7m + 8m = 2.4mV
Simplifying further:
12.7m = 2.4mV
Dividing both sides by m and substituting the value of m:
12.7 = 2.4V
Now, solving for V:
V = 12.7 / 2.4
V ≈ 5.29 m/s
Therefore, the speed of the combined mass after the trailing skater picks up the leading skater is approximately 5.29 m/s.