3.2.4 Continuous Momentum Change (F = MtΔV)

Single mass

Imagine a plasticine ball of mass m landing on the floor with a thud. Suppose the ball undergoes a velocity change of Δv during the collision, which lasted for a time duration of Δt. The impact force F can thus be calculated by

$\displaystyle F=\frac{{\Delta p}}{{\Delta t}}=m\frac{{\Delta v}}{{\Delta t}}$

Continuous stream of particles

Now imagine a heavy storm during which each raindrop undergoes the same change in velocity Δv upon landing. Since there is a continuous stream of raindrops, a constant impact force F on the floor is produced. If M kg of rain lands every t second, then the amount of momentum change per unit time would be $\displaystyle \frac{{M\Delta v}}{t}$ . If we denote the amount of rain per unit time by M_t, the formula for the impact force can be written as

$\displaystyle F=\frac{{\Delta p}}{{\Delta t}}={{M}_{t}}\Delta v$

This formula is applicable for scenarios where a stream of particles (instead of a single body) undergo momentum change. E.g. rocket fuel being ejected from the rocket, waterjet hitting a wall, etc.

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Rocket Example

A rocket ejects propellant at the rate of 300 kg s^-1 at the speed of 2000 m s^-1 (relative to the rocket). Calculate the thrust developed by the rocket at the instant when the rocket lifts off.

Solution

When 300 kg of 0 m s⁻¹ propellant becomes 300 kg of 2000 m s^-1 propellant, we are looking at a momentum gain of

$\displaystyle \begin{aligned}\Delta p&=m\Delta v\\&=300\text{ kg }\times 2000\text{ m }{{\text{s}}^{{-1}}}\\&=600,000\text{ kg m }{{\text{s}}^{{-1}}}\text{ }\end{aligned}$

This “production” of momentum requires a force. According to N2L, the rate of production of momentum is the force!

$\displaystyle \begin{aligned}F&=\frac{{\Delta p}}{{\Delta t}}\\&=600,000\text{ kg m }{{\text{s}}^{{-1}}}\div 1\text{ s }\\&=600,000\text{ kg m }{{\text{s}}^{{-2}}}\\&=600\text{ kN}\end{aligned}$

Alternatively, we can use the formula $\displaystyle F={{M}_{t}}\Delta v$ , where M_t is the rate of propellant ejection in kg s^-1.

$\displaystyle \begin{aligned}F&={{M}_{t}}\Delta v\\&=300\text{ kg }{{\text{s}}^{{-1}}}\times 2000\text{ m }{{\text{s}}^{{-1}}}\\&=600,000\text{ kg m }{{\text{s}}^{{-2}}}\text{ }\\&=600\text{ kN}\end{aligned}$

Technically, this 600 kN is the (downward) force that the rocket must exert to expel the propellant at this rate and speed. By N3L, this is also the magnitude of the (upward) thrust force experienced by the rocket.

Water Jet Example

A high-pressure water jet ejects a cylindrical column of water with cross sectional area of 2.5 cm² travelling at 80 m s^-1. The water hits a vertical wall horizontally, loses all its horizontal velocity and trickles vertically down the wall after impact. Given that the density of the water jet is 1000 kg m^-3, what is the force it exerted on the wall?

Solution

Distance travelled by water jet in 1 s $=80\text{ m}$

Volume of water hitting the wall in every 1 s $\displaystyle =(2.5\times {{10}^{{-4}}})(80)=0.020\text{ }{{\text{m}}^{3}}$

Mass of water hitting the wall in every 1 s $\displaystyle =(0.020)(1000)=20\text{ kg}$

We note that in every second, 20 kg of 80 m s⁻¹ water becomes 20 kg of 0 m s^-1 water. This “destruction” of momentum requires a force. According to N2L, the rate of “destruction” of momentum is equal to the force

$\displaystyle \begin{aligned}F&={{M}_{t}}\Delta v\\&=20\text{ kg }{{\text{s}}^{{-1}}}\times 80\text{ m }{{\text{s}}^{{-1}}}\\&=1600\text{ kg m }{{\text{s}}^{{-2}}}\text{ }\\&=1.6\text{ kN}\end{aligned}$

Technically, this 1.6 kN is the (leftward) force experience by the water jet. But by figuring out the force exerted by the wall on the water jet, we can infer the (rightward) impact force the water jet exerts on the wall. Because these two forces are an action-reaction pair.

Alternatively, we can apply the $\displaystyle F={{M}_{t}}\Delta v$ formula,

$\displaystyle \begin{aligned}(F&={{M}_{t}}\Delta v)\\F&=(\rho Av)v\\&=\rho A{{v}^{2}}\\&=(1000)(2.5\times {{10}^{{-4}}}){{(80)}^{2}}\\&=1.6\text{ kN}\end{aligned}$