Problem:

A pwrite-up of mass m1 and also velocity u1 collides through a particle of mass m2 at remainder. The two pshort articles stick together. What fraction of the original kinetic power is shed in the collision? (Simplify the answer as much as possible.)

Solution:

Concepts:Momentum conservationReasoning:In an inelastic collision kinetic power is not conserved, however momentum is conoffered.Details of the calculation:m1u1 = (m1 + m2)v.Ef = ½ (m1 + m2)v2, Ei = ½ m1u12.Fractivity of energy lost = (Ei - Ef)/Ei = 1 - m1/(m1+ m2) = m2/(m1 + m2).Problem:

A pwrite-up of mass m traveling with (non-relativistic) velocity u1 provides a head-on collision via a 2nd ppost of mass M, which is at rest in the laboratory. If the collision is entirely inelastic, what fraction of the original kinetic power continues to be after the collision?

Solution:

Concepts:Inelastic collisions, momentum conservationReasoning:In inelastic collisions: momentum is conserved, mechanical power is not conoffered.Details of the calculation:Assume the pshort article is initially traveling in the positive x-direction.Before the collision: pi = mu1, Ei = ½mu12.After the collision: pf = (m+M)v = mu1. Ef = ½(M+m)v2 = ½u12.Ef/Ei = m/(M+m).Problem:

A 36 g bullet through a rate of 350 m/s strikes a 8 cm thickfence short article. The bullet is retarded by an average force of 3.6*103 Nwhile traveling all the way with the board.(a)What speed does the bullet have actually when it emerges?(b)How many type of such boards can the bullet penetrate?

Solution:

Concepts:Newton"s second regulation, F = ma, activity via consistent accelerationin one dimensionReasoning:For activity via constant acceleration the equations of kinematicsyieldvxf2 = vxi2 + 2ax(xf- xi).Details of the calculation:(a) (xf - xi) = 0.08 m, vxf2 =(350 m/s)2 - 2*(0.08 m)(3.6*103 N)/( 36*10-3kg),vxf = 326.34 m/s.(b) vxf = 0, (350 m/s)2 =2*d(3.6*103 N)/( 36*10-3 kg), d = 0.6125 m.# of boards the bullet have the right to permeate = 7, (d/0.08 = 7.65)orWork-kinetic power theorem: d = ½mvi2.Problem:

Board A is inserted on board B as shown. Both boards slide, without relocating with respect to each other, alengthy a frictionmuch less horizontal surconfront at a rate v. Board B hits a resting board C "head-on." After the collision, boards B and also C relocate together, and board A slides on peak of board C and also stops its motion loved one to C in the position presented on the diagram. What is the length of each board? All three boards have actually the same mass, size, and also form. It is recognized tright here is no friction between boards A and also B; the coreliable ofkinetic friction in between boards A and C is μk. Solution:

Concepts:Inelastic collisions, momentum conservation, frictionReasoning:Board B and also C collide inelastically. Board A slides over boards B and C, and also we have actually sliding friction. Solve the difficulty in a inertial frame
.Details of the calculation:After the collision boards B and also C stick together and relocate with rate v/2 in the lab structure (momentum conservation). Tbelow is no friction in between A and B. Board A moves with speed v in the lab framework. The CM structure moves through velocity 2v/3 in the direction of the right. In the CM structure board A moves through velocity v/3 in the direction of the ideal and also boards B and also C relocate with velocity v/6 towards the left right after the inelastic collision. Let us work in that framework. Let the right edge of board A be at x" = 0 at t = 0, right after the collision. At time t, once the ideal edge of the board A is at x", the pressure on the board is the force of friction μN = μmg3x"/(2L) directed in the direction of the left.L is the size of each board, 3x"/(2L) represents the fraction of the height board that is supported by the surface with friction.The acceleration is a = F/m, d2x"/dt2 = -μg3x"/(2L).The solution to this differential equation is x" = x"0sinωt, ω = (3μg /(2L))½, v" = ωx"0cosωt,through the initial conditionsx"(0) = 0, v"(0) = v/3 = ωx"0.x"0 = v/(3ω) = v"(0)/(3μg/(2L))½.When x" = x"0, v" = 0. x"0 therefore is the (2/3) of size of the board, x"0 = 2L/3.2L/3 = v"(0)/(3μg/(2L))½, 2L2/9 = v"(0)2L/(3μg).The length of the board is L = 3v"(0)2/(2μg) = v2/(6μg)You deserve to also usage the work-kinetic power theorem. We have in the CM frameF = -Cx", C = 3mμg/(2L). C(2L/3)2/2 = ½m(v/3)2, L = v2/(6μg).Problem:

An hour glass sits on a range. Initially all the sand also (mass m) in the glass (mass M) is held in the upper reservoir. At t = 0, the sand is released. If it exits the top reservoir at a consistent price dm/dt = -λ, attract (and label quantitatively) a graph showing the analysis of the scale at all times t > 0.

You are watching: Kinetic energy is the force that needs to be dissipated in a collision Solution:

Concepts:Newton"s second regulation, F
= dp/dtReasoning:The reading of the scale is figured out by the amount of sand also not falling openly and the force forced to reduce the momentum of an amount of sand also dmgetting to the bottom to zero in the a time interval dt.Details of the calculation:Each grain of sand also drops a distance h. It therefore drops for a time interval t1 = (2h/g)½. The first grain hits the bottom at t = t1. For 0 1, the scale reads W1 = (M + m)g - λgt.W1 decreases lipractically from (M + m)g to (M + m)g - λ(2gh)½.At time t2 = m/λ, all the sand has actually left the upper reservoir. For t1 2, W2 = (M + m)g - λgt1 + v|dm/dt|.(M + m)g - λgt1 is the weight of the sand also that is not falling and v|dm/dt| is the pressure compelled to minimize the momentum of the amount of sand |dm| = λdt getting to the bottom in the time interval dt through speed v = (2gh)½ to zero in the time interval dt.v|dm/dt| = (2gh)½λ. λgt1 = λg(2h/g)½ = (2gh)½λ.Therefore W2 = (M + m)g. W2 is constant.At time t3 = t2 + t1 all the sand also has reached the bottom. For t2 3, W3 = W2 + λg(t - t2) .W3 boosts linearly from (M + m)g to (M + m)g + λ(2gh)½.For t > t3, W4 = (M + m)g = constant. Problem:

A 15.2 g bullet hits a 0.463 kg block from below. The initial speed of the bullet is 624 m/s and it emerges from the block at 131 m/s. (a) How high does the block rise?(b) If the block is 2.34 cm thick, estimate the average pressure on the block. Assume that the bullet passes totally via before the block moves appreciably. Solution:

Concepts:Momentum conservation, impulseReasoning:In the collision in between the bullet and the block momentum is conserved. The collision time is extremely brief.This is a 1-dimensional trouble.Details of the calculation:(a) m = 0.0152 kg , M = 0.463 kg, mvi = mvf + Mv.v = m(vi - vf)/M = 0.0152*(489 m/s)/ 0.463 = 16 m/s.The initial speed of the block is v = 16 m/s.It rises to a elevation h = v2/(2g) = 13.15 m.(b) The block receives an impulse Δp = Mv in the time interval Δt.Δt is the moment it takes the bullet to pass through the block, Δt = 2*(0.0234 m)/(vi + vf) = 6.23*10-5 s, assuming unidevelop deceleration.F = Δp/Δt = 1.19*105 N.

Collisions in 2D

Problem:

A particle of mass m strikes a frictionless, smooth, hard surchallenge at an angle θ from the surface normal. The pwrite-up bounces off via coefficient of restitution ε. (The coeffective of restitution is defined as the ratio of the magnitudes of the velocity components along the normal to the plane of call after and before the collision.) Find the rebound angle for the ppost as it leaves the surface.

Solution:

Concepts:Definition: coeffective of restitutionReasoning:This is an inelastic collision. We are asked to find the rebound angle as soon as given the coefficient of restitution.Details of the calculation:ε = |v2|/|v1|, v1 is the velocity component of the pwrite-up toward the floor prior to collision and also v2 is the velocity component of the pwrite-up away from the floor after the collision.Before the collision: tanθ = |vx|/|vy|.After the collision: tanθ" = |vx"|/|vy"| = |vx|/(ε|vy|) = (1/ε)tanθ.Since ε Problem:

A 90 kg fullback running eastern through a rate of 5 m/s is tackled by a 95 kg enemy running north with a rate of 3 m/s. If the collision is perfectly inelastic, calculate the rate and the direction of the players just after the tackle.

Solution:

Concepts:Inelastic collision, conservation of momentumReasoning:If a component of the complete external pressure acting on a system is zero, then the corresponding component of the total momentum is conserved. We assume that the horizontal component of the external force can be neglected throughout the collision.Details of the calculation:The initial momentum of player 1 is p
1= (90 kg 5 m/s)i = (450 kgm/s)i.The initial momentum of player 2 is p2 = (95 kg 3 m/s)j = (285 kgm/s)j.Momentum is conoffered, the last momentum p of both players is p = p1 + p2.p = (m1 + m2)v.v = (2.432 m/s)i + (1.54 m/s)j.v2 = 8.29(m/s)2, v = 2.88 m/s.The rate of the players after the collision is 2.88 m/s.tanθ = py/px = 285/450 = 0.63, θ = 32.34o.Their direction of take a trip renders an angle θ= 32.34o with the x-axis. (The x-axis is pointing east.)Problem:

After a fully inelastic collision between 2 objects of equal mass, each having actually initial rate v, the two relocate off along with rate v/3. What was the angle between their initial directions?

Solution:

Concepts:Conservation of momentumReasoning:The external pressure acting on the device is zero, the full momentum is conoffered.Details of the calculation: p1x + p2x = pfx = pf, p1y+ p2y = 0.2mv cosθ = 2mv/3, cosθ = (1/3), θ = 70.5o.The angle between their initial directions is 2θ = 141o.Problem:

The mass of the blue puck is 20% better than the mass of the green one. Before colliding, the pucks approach each various other with equal and opposite momenta, and also the green puck has actually an initial rate of 10 m/s. Find the speed of the pucks after the collision, if half the kinetic energy is shed in the time of the collision. Solution:

Concepts:Inelastic collisions, momentum conservationReasoning:The collision is inelastic, since energy is not conoffered.The total momentum of the 2 pucks is zero prior to the collision and also after the collision.Let particle 1 be the green puck and also ppost 2 be the blue puck. Before and also after the collision the ratio of the speeds is v2/v1= m1/m2 = 1/1.2.Details of the calculation:The final kinetic power of the device amounts to ½ times its initial kinetic energy.½m1v1i2+ ½m2v2i2 = 2(½m1v1f2+ ½m2v2f2).m1v1i2+ m2v2i2 = 2(m1v1f2+m2v2f2).m1v1i2 + 1.2m1(v1i/1.2)2= 2(m1v1f2 + 1.2m1(v1f/1.2)2).m1(v1i2 + v1i2/1.2) = 2m1(v1f2 + v1f2/1.2).v1i2 = 2v1f2, v1f = 0.707v1i.v1f = 7.07 m/s, v2f = 5.89 m/s.Problem:

A wooden block of mass M hangs on a massless rope of length L. A bullet of mass m collides with the block and remains inside the block. Find the minimum velocity of the bullet so that the block completes a full circle about the suggest of suspension. Solution:

Concepts:Inelastic collisions, conservation of momentum, circular motionReasoning:Let us treat the block/bullet combicountry as a allude pshort article. The block and bullet will certainly start moving in a circle. The pressures on the block/bullet are the pressure of gravity and also the stress in the rope.If the initial speed is big enough so that the block/bullet gains a height higher than L, then the centripetal acceleration required to store the block/bullet relocating on the circle is partly offered by gravity and the stress and anxiety decreases. Whenthe stress and anxiety vanishes then the block/bullet behaves prefer a free ppost in a gravitational area executes projectile movement.Details of the calculation:Conservation of momentum:mvbullet = (M + m)v, v = mvbullet/(M + m)When the block/bullet reaches the peak of the circle with a rate voptimal, so that the centripetal acceleration vtop2/L > g, then it will certainly completes a complete circle about the suggest of suspension. ½(M + m)v2 - 2(M + m)gL = ½(M + m)vtop2 vtop2 = v2 - 4gL = 2(vbullet)2 - 4gL.We require 2(vbullet)2/L - 4g > g --> vbullet > (5gL)½ (M + m)/m.Problem:

A vehicle of full mass M1 = M and velocity v1 renders a totally inelastic collision at time t = 0 with a second auto of mass M2= 2M at remainder. Before the collision a point object of mass m 1 will certainly the little mass lose call through the surchallenge of the cavity?

Solution:

Concepts:Inelastic collisions, Newton"s regulations, conservation of momentum, circular motion, frame transformationsReasoning:Before the collision automobile 1 and also mass m move with unidevelop velocity v1. The impulse of the collision changes the velocity of car 1, and also after the collision automobile 1 moves via uniform velocity v2. The tiny mass moves through velocity v = v1 - v2 through respect to automobile 1 instantly after the collision. After the collision the remainder structure of car 1 is an inertial framework and we have the right to analyze the difficulty in that frame.Details of the calculation:Conservation of momentum: Mv1 = 3Mv2, v2 = v1/3, v = v1 - v2 = 2v1/3.In the rest structure of vehicle 1 automatically after the collision we have actually a particle of mass m at the bottom of a cavity of radius r moving with velocity v. The pshort article will begin relocating in a circle. The forces on the ppost are the force of gravity and also the normal force from the wall of the cavity. If the initial speed is huge sufficient so that the ppost gains a height greater than r, then the centripetal acceleration necessary to store the pwrite-up relocating on the circle is partly offered by gravity and the normal force decreases. Whenthe normal force vanishes then the ppost behaves like a complimentary particle in a gravitational field and looses contact via the wall. When ½mv2 ½, or v1 ½ the pshort article will not break call with the wall.When the pshort article reaches the height of the cavity with a speed vheight, so that the centripetal acceleration vtop2/r > g, then it will not break call through the wall. Then ½mv2 = 2mgr + ½mvtop2 > 2mgr + ½mgr = (5/2)mgr.v > (5gr)½, v1 > (3/2)(5gr)½.For initial velocities (3/2)(2gr)½ 1 ½ the particle breaks contact via the wall.

See more: How Far Can The Average Person Swim Without Stopping? How Far Could A Human Swim

Problem:

A brick is thrown (from ground level) via rate V at an angle θ with respect to the (horizontal) ground. Assume that the long confront of the brick remains parallel to the ground at all times, and that there is no dedevelopment in the ground or the brick once the brick hits the ground. If the coeffective of friction between the brick and also the ground is μ what should θ be so that the brick travels the maximum complete horizontal distance before lastly coming to rest?

Solution:

Concepts:Projectile activity, impulsive forcesReasoning:The normal force from the ground has to administer an impulse to reduce the vertical component of the momentum of the brick to zero.Details of the calculation:Let V be the initial rate. The horizontal speed and also initial vertical speed are then Vcosθ and Vsinθ, respectively. The horizontal distance traveled in the air is the standarddair = 2V2sinθcosθ/g.To discover the distance traveled alengthy the ground, we have to determine the horizontal speed just after the affect has emerged. The normal force, N, from the ground is what reduces the vertical rate from V sinθ to zero, during the impact. So we have ∫ N dt = m V sinθ,where the integral runs over the moment of the influence. But this normal force (when multiplied by μ to provide the horizontal friction force) additionally produces a sudden decrease in the horizontal rate, in the time of the time of the influence. So we have m ∆vx = -∫ μN dt = -μ m Vsinθ,which implies ∆vx = -μ Vsinθ.(We have actually neglected the result of the mg gravitational pressure throughout the short time of the affect, since it is much smaller sized than the N impulsive force.) As such, the brick begins its sliding motion through speedv = Vcosθ - μ Vsinθ.(Keep in mind that this is true just if tanθ The friction force from this suggest on is μmg, so the acceleration is a = -μg. The distance traveled along the ground isdground = (Vcosθ - μ Vsinθ)2/(2μg).We desire to uncover the angle that maximizes the total distance, dtotal = dair + dground.d(dtotal)/dθ = 0. d(sinθcosθ + cos2θ/(2μ) + μ sin2θ/2)/dθ = 0.cos2θ - sin2θ + sinθ cosθ (μ - 1/μ) = 0. cos(2θ) + sin(2θ)½(μ - 1/μ) = 0.tan(2θ) = 2μ/(1 - μ2). Using the trigonometric identity, tan(2θ) = 2 tanθ/(1 - tan2θ) we have actually tanθ = μ.Taking the derivative via respect to θ, we view that the maximum total distance is achieved when tanθ = μ. Note, however, that the over analysis is valid just if tanθ 1, then the optimal angle is θ = 45o. (The brick stops after the impact, and also μ = 45o gives the maximum value for the dair.Problem:

A marble bounces down stairs in a continuous manner, hitting each step at the exact same area and also bouncing the same elevation over each step. The stair height equals its depth (tread = rise) and also the coefficient of restitution ε is provided. Find the essential horizontal velocity and also bounce height. (The coreliable of restitution is characterized as ε = -vf/vi, wbelow vf and vi are the vertical velocities just after and before the bounce. respectively).