SCIENCE 9 Unit 4, Module 1 Activity 5 Conservation of Momentum Direction: Solve the following problems systematically. 1. A \( 1600-\mathrm{kg} \) car is moving at \( 35 \mathrm{~m} / \mathrm{s} \) northward. A cement delivery truck with a mass of \( 2500 \mathrm{~kg} \) moving at \( 20 \mathrm{~m} / \mathrm{s} \) southward collides head-on with the car. If they move together after collision, which vehicle will push another? Why? In what direction? 2. Bennie is at the carnival playing some of the arcades. In one booth he throws a \( 0.40 \mathrm{~kg} \) ball forward at \( 21 \mathrm{~m} / \mathrm{s} \) in order to hit a \( 0.20 \mathrm{~kg} \) bottle which is on the table. The \( 0.20-\mathrm{kg} \) bottle goes flying forward at \( 30 \mathrm{~m} / \mathrm{s} \) when hit. What is the velocity of the ball after it hits the bottle?
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1. An arrow with a mass of 115 g and a velocity of 50.0 m/s hits and sticks to a piece of wood of mass 2.10 kg. If the block of wood is on a frictionless, horizontal surface, with what velocity will the block move? [2 marks] 2. A 2.50 kg mass moving 7.50 m/s to the right collides head on with a 4.90 kg mass. After the collision the 2.50 kg mass is moving 5.00 m/s to the left and the 4.90 kg mass is moving 4.88 m/s to the right. [ 2 marks] a. Calculate the velocity of the 4.90 kg mass before the collision b. If the collision lasts for 0.0625 seconds calculate the force that acted on each mass during the collision. 3. An 8.00 kg shell initially moving at 28.8 m/s [East] explodes into 3 fragments. A 3.00 kg fragment flies directly north at 50.0 m/s and a 2.00 kg fragment flies directly east at 40.0 m/s. Calculate the velocity of the third fragment. [4 marks]
Aishwarya K.
Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. (a) Is it possible that the two-object system has a final total momentum of zero after the collision? (b) Roughly, what is the direction of the final total momentum of the two-object system after the collision? Problem Object A has a mass of $m_{A}=17.0 \mathrm{~kg}$ and an initial velocity of $\overrightarrow{\mathbf{v}}_{0 \mathrm{~A}}=8.00 \mathrm{~m} / \mathrm{s},$ due east. Object $\mathrm{B},$ however, has a mass of $m_{\mathrm{B}}=29.0 \mathrm{~kg}$ and $\mathrm{an}$ initial velocity of $\overrightarrow{\mathbf{v}}_{0 \mathrm{~B}}=5.00 \mathrm{~m} / \mathrm{s},$ due north. Find the magnitude and direction of the total momentum of the two-object system after the collision. Make sure that your answers are consistent with your answers to the Concept Questions.
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
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