The change in momentum (Δp) is expressed in kg m/s (kilogram meters per second). The mass is expressed in kg. The change in velocity is expressed in m/s (meters per second). Δv can also be expressed as vf - vi where vf = the object’s final velocity and vi = the object’s initial (or starting) velocity. Therefore, you might see the equation Δp = m(Δv) represented as Δp = m(vf - vi).
The change in momentum (Δp) is expressed in kg m/s (kilogram meters per second). The force (F) is expressed in Newtons. In most cases, Newtons are abbreviated as “N”. The amount of time the force was applied (Δt) is expressed in seconds. Δt is also called the “time interval. ”
We knew to use the formula Δp =m(Δv) because we were provided with mass (in kg) and velocities (in m/s). Remember that the change in momentum (Δp) is always expressed in kg m/s.
We knew to use the formula Δp = F(Δt) because we were provided with the Force (in N) and time (in seconds). The change in momentum (Δp) is always expressed in kg m/s, regardless of which formula you use.
Since we were provided with mass and velocity, we should use the formula Δp = m(Δv). We know that m = 55kg. Because the sled was stationary when it began moving, we know that it accelerated from 0 m/s to 11 m/s. This means that Δv = 11 m/s - 0 m/s. Therefore, Δv = 11 m/s. We can now plug in the values for m and Δv into the formula: Δp = 55 kg * 11 m/s, which equals 605 kg m/s. Therefore, the sled’s change in momentum is 605 kg m/s.
Since we’re provided with a force and a time, we should use the formula Δp = F(Δt). We know the force was 3,000,000 N. Because the rocket engine only applies a force while it’s burning, we know that the force was applied to the spacecraft for a total of 45 of seconds. We can now plug in the values for F and Δt into the formula: Δp = 3,000,000 N * 45 s, which equals 135,000,000 kg m/s. Therefore, the change in the rocket’s momentum is 135,000,000 kg m/s (or 1. 35*10^8 kg m/s).
Because we have mass and velocity, we should use Δp =m(Δv). We know the mass is 14,000 kg. Because the plane accelerated from 200 m/s to 550 m/s, we know that Δv = 550 m/s - 200 m/s. Therefore, Δv = 350 m/s. In other words, the aircraft’s change in velocity is 350 m/s. Plugging in these values gives us Δp = 14,000 kg * 350 m/s, which equals 4,900,000 kg m/s. Therefore, the aircraft’s change in momentum is 4,900,000 kg m/s.
Because we’re provided with a force and a time, we should use the formula Δp = F(Δt). We know F = 400 N. We know that the scientist observed the boat for a total of 60 s. However, the boat’s engine did not begin applying a force until 25 s had passed. Therefore, to find the time that the force was actually applied (Δt), we need to find the difference between 60 s and 25 s. Δt = 60 s - 25 s. Therefore, Δt = 35 s. In other words, the force was applied to the boat for a total of 35 seconds. Plugging in the values for F and Δt into our formula, Δp = F(Δt), we know that Δp = 400 N * 35 s, which equals 14,000 kg m/s. Therefore, the boat’s change in momentum is 14,000 kg m/s.
