A 58 kg man is riding on a 60 kg cart traveling at a speed of 1 m/s. He jumps off with zero horizontal speed (relative to the ground). What is the resulting change in the speed of the cart?

To solve this problem, we need to use the principle of conservation of momentum. The total momentum before the man jumps off is equal to the total momentum after he jumps off.

Step 1: Calculate the initial momentum The initial momentum is the mass of the system (man + cart) times the velocity of the system. The mass of the system is 58 kg (man) + 60 kg (cart) = 118 kg. The velocity of the system is 1 m/s. So, the initial momentum is 118 kg * 1 m/s = 118 kg*m/s.

Step 2: Calculate the final momentum After the man jumps off, the cart is the only thing moving. The man's speed relative to the ground is zero, so he does not contribute to the final momentum. The final momentum is therefore the mass of the cart times its final velocity. We'll call the final velocity of the cart v. So, the final momentum is 60 kg * v.

Step 3: Set the initial momentum equal to the final momentum According to the principle of conservation of momentum, the initial momentum is equal to the final momentum. So, we have 118 kg*m/s = 60 kg * v.

Step 4: Solve for v To solve for v, we divide both sides of the equation by 60 kg. This gives us v = 118 kg*m/s / 60 kg = 1.97 m/s.

Step 5: Calculate the change in speed The change in speed of the cart is the final speed minus the initial speed. So, the change in speed is 1.97 m/s - 1 m/s = 0.97 m/s.

So, the resulting change in the speed of the cart is 0.97 m/s.