Consider a stationary skater of mass 60 kg. Another skater of mass 40kg rushes and holds the first at a speed of 12km/hr. Both skaters hold together and move as one system (S) with a speed V.
a) Determine V.
b) Determine the kinetic energy of (S) before and after collision.
conservation of momentum:
60(0)+40(12)=(60+40)V
solve for b.
Ke before= 1/2 40*12^2
KE after= 1/2 (100)V^2
Hoda
To solve this problem, we can use the principle of conservation of momentum and conservation of kinetic energy.
a) Determine V:
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the first skater is given by:
p1 = m1 * v1
where m1 is the mass of the first skater (60 kg) and v1 is the initial velocity (0 m/s) since the skater is stationary.
The initial momentum of the second skater is given by:
p2 = m2 * v2
where m2 is the mass of the second skater (40 kg) and v2 is the initial velocity (12 km/hr).
We first need to convert v2 from km/hr to m/s:
v2 = 12 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 3.33 m/s
The total initial momentum before the collision is:
p_initial = p1 + p2 = (60 kg * 0 m/s) + (40 kg * 3.33 m/s)
The final momentum after the collision is given by:
p_final = (m1 + m2) * V
Using the principle of conservation of momentum:
p_initial = p_final
(40 kg * 3.33 m/s) + (60 kg * 0 m/s) = (60 kg + 40 kg) * V
Simplifying the equation:
133.2 kg m/s = 100 kg * V
Divide both sides by 100 kg to find V:
V = 133.2 kg m/s / 100 kg = 1.33 m/s
Therefore, the speed of the system (S) after the collision is 1.33 m/s.
b) Determine the kinetic energy of (S) before and after collision:
The initial kinetic energy (KE_initial) of the system is the sum of the individual kinetic energies before the collision.
KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
= (1/2) * 60 kg * (0 m/s)^2 + (1/2) * 40 kg * (3.33 m/s)^2
Simplifying the equation:
KE_initial = 0 J + 222.4 J = 222.4 J
The final kinetic energy (KE_final) of the system is the sum of the individual kinetic energies after the collision.
KE_final = (1/2) * (m1 + m2) * V^2
= (1/2) * 100 kg * (1.33 m/s)^2
Simplifying the equation:
KE_final = 88.55 J
Therefore, the kinetic energy of the system (S) before the collision is 222.4 J and after the collision is 88.55 J.
To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy principles.
a) To determine V, we need to apply the conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
Initially, the stationary skater has no momentum (since it is not moving), so the total momentum before the collision is zero. The second skater, with a mass of 40 kg, is moving with a speed of 12 km/hr. We need to convert this speed to m/s:
Speed = 12 km/hr = (12 * 1000) m / (3600) s = 33.33 m/s
Therefore, the momentum of the second skater can be calculated as:
Momentum = mass × velocity = 40 kg × 33.33 m/s = 1333.2 kg·m/s
After the collision, both skaters hold together and move as one system (S) with a speed V. The total mass of the system is 60 kg + 40 kg = 100 kg.
So, the total momentum after the collision can be given as:
Total momentum = mass of system × velocity of system = 100 kg × V
Applying the conservation of momentum principle, we can write the equation:
Total momentum before the collision = Total momentum after the collision
0 = 100 kg × V
Solving for V:
V = 0 m/s
Therefore, the speed of the system after the collision is zero. This means the skaters come to a complete stop.
b) To determine the kinetic energy of the system before and after the collision, we can use the equation:
Kinetic energy = 0.5 × mass × velocity^2
Before the collision:
The initial kinetic energy of the system is given by the sum of the individual kinetic energies of the skaters:
Kinetic energy before = 0.5 × mass1 × velocity1^2 + 0.5 × mass2 × velocity2^2
Since the first skater is stationary (velocity1 = 0), their kinetic energy is zero. Therefore:
Kinetic energy before = 0.5 × 40 kg × (33.33 m/s)^2 = 22,220 J
After the collision:
Since the system comes to a stop (V = 0), the final kinetic energy of the system is also zero.
Kinetic energy after = 0 J
Therefore, the kinetic energy of the system before the collision is 22,220 J, and after the collision, it becomes 0 J.