Learning Materials
Features
Discover
Chapter 8: Problem 21
A ball having a mass of \(0.500 \mathrm{~kg}\) is thrown at a speed of \(20 \mathrm{~m} / \mathrm{s}\). Determine the magnitude of its momentum.
Over 30 million students worldwide already upgrade their learning with 91影视!
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?
A rocket standing on its launch platform points straight upward. Its engines are activated and eject gas at a rate of \(1500 \mathrm{~kg} / \mathrm{s}\). The molecules are expelled with an average speed of \(50 \mathrm{~km} / \mathrm{s}\). How much mass can the rocket initially have if it is slow to rise because of the thrust of the engines? The problem provides mass flow and speed, the product of which is equivalent to the time rate-of-change of momentum. That should bring to mind the impulse- momentum relationship, which, of course, is Newton's Second Law. Since the initial motion of the rocket itself is negligible in comparison to the speed of the expelled gas, we can assume the gas is accelerated from rest to a speed of \(50 \mathrm{~km} / \mathrm{s}\). The impulse required to provide this acceleration to a mass \(m\) of gas is from which $$ \begin{array}{c} F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 \\ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{\mathrm{m}}{\mathrm{L}} \end{array} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is 1500 \(\mathrm{kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is 1500 \(\mathrm{kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ An equal but opposite reaction force acts on the rocket, and this is the upward thrust on the rocket. The engines can therefore support a weight of \(75 \mathrm{MN}\), so the maximum mass the rocket could have is $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$
Two balls of equal mass, moving with speeds of \(3 \mathrm{~m} / \mathrm{s}\), collide head-on. Find the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(1 / 3\).
Three point masses are placed on the \(x\) -axis: \(200 \mathrm{~g}\) at \(x=0,500 \mathrm{~g}\) at \(x=30 \mathrm{~cm}\), and \(400 \mathrm{~g}\) at \(x=70 \mathrm{~cm}\). Find their center of mass. We can make the calculation with respect to any point, but since all the data is measured from the \(x=0\) origin, that point will do nicely. $$ x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(0.20 \mathrm{~kg})+(0.30 \mathrm{~m})(0.50 \mathrm{~kg})+(0.70 \mathrm{~m})(0.40 \mathrm{~kg})}{(0.20+0.50+0.40) \mathrm{kg}}=0.39 \mathrm{~m} $$ The center of mass is located at a distance of \(0.39 \mathrm{~m}\), in the positive \(x\) -direction, from the origin. The \(y\) - and z-coordinates of the center of mass are zero.
A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a \(4.0\) -g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide headon and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which KE is not conserved, although momentum is. Let the 16 -g mass be \(m_{1}\) and the 4.0-g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the \(4.0\) -g mass has a scalar value of \(u_{2 x}\) \(=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: $$ \begin{array}{l} \text { Momentum before impact = Momentum after impact }\\\ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $$ (Notice that the \(4.0\) -g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=\) \(0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine