Take rightward as positive. A 1 kg puck at +4 m/s hits a parked 2 kg puck, and the pucks bounce apart. Afterward the 1 kg puck moves at −0.5 m/s. Enter the 2 kg puck's velocity after the collision in m/s (include the sign).
On a level track, a 3 kg cart rolling at 4 m/s catches a 1 kg cart rolling at 2 m/s the same way, and they couple.
The pair rolls on at 3.5 m/s.
How much kinetic energy did the coupling convert to other forms?
Count the kinetic energy before. The 3 kg cart carries (0.5 × 3) × (4 × 4) = 24 J.
The 1 kg cart carries (0.5 × 1) × (2 × 2) = 2 J. Together: 24 + 2 = 26 J before the coupling.
Count the kinetic energy after. One body of 3 kg + 1 kg = 4 kg moves at 3.5 m/s: (0.5 × 4) × (3.5 × 3.5) = 24.5 J.
Subtract: 26 - 24.5 = 1.5. The coupling converted 1.5 J of kinetic energy to thermal energy and permanent deformation at the latch.
Both totals count every mover: each cart's kinetic energy before, the coupled body's after.
The coupled pair's speed is the collision's own outcome; with it in hand, the energy count is two totals and one subtraction.
A 3 kg cart at 4 m/s catches a 1 kg cart at 2 m/s moving the same way; they couple and roll on at 3.5 m/s. Calculate the kinetic energy converted to other forms.
Write down the values in the question:
m1 = 3 kg at 4 m/s; m2 = 1 kg at 2 m/s; coupled: 4 kg at 3.5 m/s
Total the kinetic energy before:
(0.5 × 3) × (4 × 4) = 24
(0.5 × 1) × (2 × 2) = 2
24 + 2 = 26. KE before = 26 J.
Total the kinetic energy after:
(0.5 × 4) × (3.5 × 3.5) = 24.5. KE after = 24.5 J.
Subtract:
26 - 24.5 = 1.5
The coupling converted 1.5 J of kinetic energy to thermal energy and permanent deformation.
A 4 kg mail cart rolling at 3 m/s catches a 1 kg mail cart rolling at 2 m/s the same way on a level hallway. Enter the total kinetic energy of the two carts just before they couple, in J.
A 4 kg mail cart and a 1 kg mail cart have just coupled and roll on together at 2.8 m/s. Enter the kinetic energy of the coupled pair, in J.
A 1.5 kg wagon at 6 m/s catches a 4.5 kg wagon at 2 m/s on the same line; they latch and move at 3 m/s. Calculate the kinetic energy converted.
Write down the values in the question:
m1 = 1.5 kg at 6 m/s; m2 = 4.5 kg at 2 m/s; coupled: 6 kg at 3 m/s
Total the kinetic energy before:
(0.5 × 1.5) × (6 × 6) = 27
(0.5 × 4.5) × (2 × 2) = 9
27 + 9 = 36. KE before = 36 J.
Total the kinetic energy after:
(0.5 × 6) × (3 × 3) = 27. KE after = 27 J.
Subtract:
36 - 27 = 9
The latch converted 9 J of kinetic energy to thermal energy and permanent deformation.
A 5 kg cart rolling at 4 m/s catches a 1 kg cart rolling at 1 m/s the same way; they couple and roll on at 3.5 m/s. Enter the kinetic energy converted to other forms, in J.
A 6 kg mine car rolling at 5 m/s catches a 4 kg mine car rolling at 2.5 m/s on the same straight rail; they couple and roll on at 4 m/s. Enter the kinetic energy converted in the coupling, in J.
A 1 kg glider moving at 12 m/s catches a 4 kg glider moving at 2 m/s on a long air track; they latch and move on at 4 m/s. Enter the kinetic energy converted in the latch, in J.
A 5 kg lab cart moving at 6 m/s catches a 3 kg lab cart moving at 2 m/s on the same track, and they couple. Enter the kinetic energy converted in the coupling, in J.