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Answer Happy • 14. (20 points) A ballistic pendulum consists of a stationary wooden block of mass M-4.92 kg suspended from a string of
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14. (20 points) A ballistic pendulum consists of a stationary wooden block of mass M-4.92 kg suspended from a string of

Posted: Sat Jul 02, 2022 10:17 pm
by answerhappygod
14 20 Points A Ballistic Pendulum Consists Of A Stationary Wooden Block Of Mass M 4 92 Kg Suspended From A String Of 1
14 20 Points A Ballistic Pendulum Consists Of A Stationary Wooden Block Of Mass M 4 92 Kg Suspended From A String Of 1 (55.6 KiB) Viewed 12 times
14 20 Points A Ballistic Pendulum Consists Of A Stationary Wooden Block Of Mass M 4 92 Kg Suspended From A String Of 2
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14 20 Points A Ballistic Pendulum Consists Of A Stationary Wooden Block Of Mass M 4 92 Kg Suspended From A String Of 3
14 20 Points A Ballistic Pendulum Consists Of A Stationary Wooden Block Of Mass M 4 92 Kg Suspended From A String Of 3 (61.73 KiB) Viewed 12 times
14. (20 points) A ballistic pendulum consists of a stationary wooden block of mass M-4.92 kg suspended from a string of vanishingly small mass. A bullet of mass m = 0.08 kg, traveling at a speed v; = 350 m/s, strikes the wooden block of the ballistic pendulum. The block (with the bullet in it) swings to a maximum height y above the initial position (see Figure 11(B)). Include units in all your numerical responses!! 111 3D D-350 m/s (A) M m+M V=7 (B) yf=? Figure 11 (a) Determine the linear momentum and kinetic energy of the bullet before it strikes the wood block. Refer Figure 11(A) Please show all your work. In other words, NO MEMORIZED EQUATIONS PLEASE!!! (b) Determine speed (p) of the block and the bullet immediately after the collision (immediately after the bullet is embedded in the block)? Refer Figure 10(B). Please show all your work. In other words, NO MEMORIZED EQUATIONS PLEASE!!!
Work and Energy • Work done by a constant external force F exerted on a system while that system undergoes a displacement d is W=Fd cos@ where F is the magnitude of the force, d is the magnitude of the displacement, and is the angle between the vector F and the vector d. • Kinetic energy: K= Zmo² • Gravitational potential energy: Ug = mgy, Elastic potential energy: U, = kx², where x is the displacement of the spring from its equilibrium (or unstretched) position (x = 0). Work-Energy equation states that K+U+U+AE = K₁+U₂+U+W where W is the work done by all the external forces acting on the system and AEth is the change in thermal energy of the system. • The increase in thermal energy of the system due to friction is AE = Wyl, where W/l is the absolute value of the work done by the force of kinetic friction. Rotational dynamics →The torque exerted by a force F: T=r_F, where r is the moment arm (or lever arm). →For an object in equilibrium: Er=0 and EF-0 → The moment of inertia of a system of particles: I = Emr² = myr + m₂r] + mər} + ... The moment of inertia of various objects: →For a single particle of mass m in circular motion of radius r: Imrª² →For a solid disk of mass M and radius R rotating about an axis passing through its center: IM MR² →For a hoop (or ring) of mass M and radius R rotating about an axis passing through its center: I MR →Newton's second law for rotational motion: Erla, where I is the moment of inertia of the rotating object Angular momentum or rotational momentum: L-lw →If the net torque acting on a rotating rigid body is zero, i.e., Er0, the the angular momentum of the rigid body does not change, i.e., L.- la constant or ha la
Relevant equations & conversions Symbols have their usual meanings 1 revolution=2r radians 1 minute = 60 seconds; 1 hour = 60 minutes • 1 kg = 1000 g: 1 mi=1609 m; 1 m-3.28 ft; 1 m = 100 cm • Magnitude of the acceleration due to gravity: g = 9.8 m/s². Relevant equations • Roots of the quadratic equation Ax² + Bx+C =0: x=- 24 Area of a triangle of base band height : A = bh • Area of a rectangle of length I and width : A=Iw. • Circumference of a circle of radius r. C = 2mr. Area of a circular disk of radius r. A = ² Trigonometric functions Newton's second law of motion [2F₂ – ΣΕ = md OR ma -B+√B²-4AC = ------* 4-4-4 Figure 1 • Weight of an object of mas mmg (magnitude). • Hooke's law for the restoring force exerted by a spring): (Fp)-kAr, where k is the spring constant and Ax is the displacement from the equilibrium position. • Frictional forces →Magnitude of the maximum force of static friction: f, mas M., where is the magnitude of the normal force acting on the body and μ, is the coefficient of static friction →Magnitude of the force of kinetic friction: fi-in, where pa is the coefficient of kinetic friction Uniform circular motion: The acceleration of an object moving in a circle of radius r at a constant speed pointing toward the center of the circle Impulse and momentum m Linear momentum of an object of mass m and velocity Impulse of a force - area under the Force-Time curve F At Impulse-momentum principle (or Newton's second law) Ap- Tor Af-LF At → Law of conservation of momentum The total momentum of a system P-F+B+ an isolated system is a constant Le P