4. Two objects collide head-on and stick together. One object had mass m and velocity v and the other had mass 2m and velocity -2u. (a) What is the velocity of the combined object? (b) How much energy was lost in the collision? (Express in terms of m and v.)
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- Show that the kinetic energy of a particle of mass m is related to the momentum p of that particleby, ? =p^2/2m. (Hint: plug mv in for p and simplify!)3 - A bomb of mass M with initial velocity vo and suddenly explodes to two fragments of mass M/3 and 2M/3 as shown in the figure. The explosion was not elastic and the gain in the energy equals Q = -SMvz. Find the velocities of the two fragments after the explosion. [Assume that vo, vị and vz all on the same axis] (1.5) V2 2M MA projectile has a mass of 10g and speed of 0.6c strikes a wall. What is the momentum of the projectile and what is the kinetic energyof the projectile? If the collision time for each projectile is 10 milliseconds what is the magnitude of the average force on the wall due to a single projectile?
- 3. Consider the following two types of completely inelastic collisions between Турe A two identical masses, m. Type A: a m m particle of mass m and velocity v strikes V a stationary target consisting of an identical particle of mass m. Type B: Туре В each particle has a mass m and a speed m m v/(22) – in this case they collide head v/2/2 V/22 on. Ignore gravity, and express your answers in terms of m and v. A. What is the speed of the center of mass for type A, and for type B collisions (before the collision occurs)? What the center of mass speed after the collision for each type of collision. Justify your answers. В. ( What is the change in the total kinetic energy for type A and type B ?A 1400 kg car at rest at a stop sign is rear ended by a 1860 kg truck traveling at a speed of 19.1 m/s. After the collision, the two vehicles are locked together. Alice is moving past the collision site in a frame S' that is traveling at a constant velocity of 6.88 m/s in the direction of travel of the incident truck. (a) Determine the momentum of the vehicles before the collision Alice's frame. kg - m/s (b) Determine the momentum of the vehicles after the collision Alice's frame. kg - m/s (c) Does Alice conclude that momentum is conserved during the collision? O Ves O No O cannot be determinedA radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of pe = 8.95×10-23 kg-m/s and Py = 6.18x10-23 kg. m/s, respectively. (Figure 1) Figure Pnuc (nuc) +y 0 Onuc Pv Pe Part A Determine the magnitude of the momentum of the second (recoiling) nucleus. Express your answer to three significant figures and include the appropriate units. Pnuc = Submit Part B ■ μÅ Value Request Answer Pearson Units Review | Constan ?
- There are two relativistic particles of identical mass m that move towards each other along the same axis. The magnitudes of their momenta are the same. They collide head-on to form a new particle of mass 8m. What were the magnitudes of momenta of the initial particles?2. A projectile of mass m = target of mass M = 2 kg and gets imbedded in it. The energy lost (AE) during this 0.2kg and initial speed v; = 200 ms-1 hits a stationary collision is closest to (a) AE = 0J (no energy is lost) (b) AE = -3640 J (c) AE = -3670 J (d) ДЕ - = -4000 JNone
- If 2 marbles (of same mass and speed) travel from one end of a straight line runway and strike a line of marbles, is it possible to have only one marble to pop off the other end of the line at twice the speed? Why? (explain based on momentum and kinetic energy) I tried to find the momentum for the single marble and got 2mv and found the kinetic energy to be 2mv^2 but I'm not sure if it's right or how to how to explain it if it is possible or not.A saturn 5 rocket must travel 17, 800 mph to escape the gravitational pull of the earth and enter space. The rocket weight 6.2 million pounds, and the initial specific impulse lasts 263 seconds. What impulse is required for the rocket to meet the 17, 800 mph momentum in needs, using the formula FT=MV?Question 2