Two 21.0 kg ice sleds are placed a short distance apart, one directly behind the other, as shown in the figure. A 3.42 kg cat, initially standing on sled 1, jumps across to sled 2 and then jumps back to sled 1. Both jumps are made at a horizontal speed of 2.84 m/s relative to the ice.
What is the final speed of sled 2? (Assume the ice is frictionless.)
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after any interaction or collision.
Given:
Mass of sled 1 (m1) = 21.0 kg
Mass of sled 2 (m2) = 21.0 kg
Mass of the cat (mc) = 3.42 kg
Velocity of the cat (vc) = 2.84 m/s
Step 1: Calculate the initial momentum of the system.
Before the cat jumps, the sleds are stationary, so the initial velocity of both sleds is zero.
The initial momentum of sled 1 (p1_initial) = m1 * v1_initial = 21.0 kg * 0 m/s = 0 kg m/s
The initial momentum of sled 2 (p2_initial) = m2 * v2_initial = 21.0 kg * 0 m/s = 0 kg m/s
The total initial momentum (p_initial) = p1_initial + p2_initial = 0 kg m/s + 0 kg m/s = 0 kg m/s
Step 2: Calculate the final momentum of the system.
After the cat jumps, it imparts a momentum to both sleds.
The final momentum of sled 1 (p1_final) = m1 * v1_final
The final momentum of sled 2 (p2_final) = m2 * v2_final
The total final momentum (p_final) = p1_final + p2_final
Step 3: Apply the conservation of momentum principle.
According to the principle of conservation of momentum, the total final momentum (p_final) should be equal to the total initial momentum (p_initial).
p_final = p_initial
p1_final + p2_final = 0 kg m/s
Step 4: Apply the principle of conservation of momentum to find the final speed of sled 2.
Since the cat jumps from sled 1 to sled 2 and then back to sled 1 without any external forces acting on the system, the total momentum remains constant.
The momentum the cat imparts to sled 1 is equal in magnitude but opposite in direction to the momentum it imparts to sled 2. This means that the final momentum of sled 1 and the final momentum of sled 2 will cancel each other, resulting in a total final momentum of zero.
p1_final + p2_final = 0 kg m/s
m1 * v1_final + m2 * v2_final = 0 kg m/s
Let's substitute the values:
21.0 kg * v1_final + 21.0 kg * v2_final = 0 kg m/s
Step 5: Solve the equation for the final velocity of sled 2 (v2_final).
Given that the cat's velocity is 2.84 m/s relative to the ice, it imparts an equal and opposite change in momentum to both sleds. Therefore, the final velocity of sled 2 (v2_final) will also be 2.84 m/s.
21.0 kg * v1_final + 21.0 kg * 2.84 m/s = 0 kg m/s
Now let's solve for v1_final:
v1_final = (21.0 kg * - 2.84 m/s) / 21.0 kg
v1_final ≈ -2.84 m/s
The negative sign indicates that sled 1 moves in the opposite direction to the velocity of the cat when it jumps.
Therefore, the final speed of sled 2 is also 2.84 m/s.
To determine the final speed of sled 2, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the jumps is equal to the total momentum after the jumps in the absence of external forces.
Before the jumps, the total momentum is given by the sum of the individual momenta of the cat and the sleds. Since the cat is initially standing on sled 1, its momentum is zero (as its initial velocity is zero). The momentum of sled 1 is also zero since it is initially at rest. Therefore, only sled 2 contributes to the initial total momentum.
After the jumps, the cat has moved to sled 2 and then jumps back to sled 1. In both cases, the cat moves horizontally at a speed of 2.84 m/s relative to the ice. Each jump involves a change in the direction of the cat's velocity, but the magnitude of the velocity remains the same. Therefore, there is no change in the cat's momentum during the jumps.
Since the ice is frictionless, there are no external forces acting on the system, so the total momentum before the jumps is equal to the total momentum after the jumps. Let's denote the final speed of sled 2 as Vf.
The initial total momentum is given by the momentum of sled 2:
Momentum_initial = mass_sled_2 * velocity_initial_sled_2
And the final total momentum is given by the momentum of sled 1 and sled 2 after the cat jumps back to sled 1 (since the cat's momentum doesn't change):
Momentum_final = (mass_sled_1 + mass_sled_2) * Vf
Setting the initial momentum equal to the final momentum:
mass_sled_2 * velocity_initial_sled_2 = (mass_sled_1 + mass_sled_2) * Vf
Rearranging the equation to solve for Vf:
Vf = (mass_sled_2 * velocity_initial_sled_2) / (mass_sled_1 + mass_sled_2)
Now we can substitute the given values:
mass_sled_1 = mass_sled_2 = 21.0 kg (since both sleds have the same mass)
mass_cat = 3.42 kg
velocity_initial_sled_2 = 2.84 m/s
Therefore,
Vf = (21.0 kg * 2.84 m/s) / (21.0 kg + 21.0 kg + 3.42 kg)
Calculating the final speed of sled 2 will give us the answer to the question.
First find the velocity of sled 1 (V):
p1=p2
0=mv+MV(1)
V(1)=-mv/M
For sled 2:
0+mv=(m+M)V
V=mv/(m+M)
When the cat leaves again:
(m+M)V=MV(2)-mv
V(2)=((m+M)V+mv)/M
(V comes from the equation above)
For sled 1:
MV(1)+mv=(M+m)V
V(1) from the first equation