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Momentum — Diagnostic Tests

Unit Tests​

UT-1: Two-Dimensional Collision​

Question:

Particle AA of mass 2.0 kg2.0\,\text{kg} moves with velocity (4.0i^+3.0j^) m s−1(4.0\hat{i} + 3.0\hat{j})\,\text{m}\,\text{s}^{-1} on a smooth horizontal surface. It collides with stationary particle BB of mass 3.0 kg3.0\,\text{kg}. After the collision, AA moves with velocity (−1.0i^+4.0j^) m s−1(-1.0\hat{i} + 4.0\hat{j})\,\text{m}\,\text{s}^{-1}.

(a) Calculate the velocity of BB after the collision.

(b) Determine whether the collision is elastic.

(c) Calculate the impulse exerted on AA during the collision.

Solution:

(a) By conservation of momentum (vector form):

mAuA+mBuB=mAvA+mBvBm_A\mathbf{u}_A + m_B\mathbf{u}_B = m_A\mathbf{v}_A + m_B\mathbf{v}_B

2.0(4.0i^+3.0j^)+0=2.0(−1.0i^+4.0j^)+3.0vB2.0(4.0\hat{i} + 3.0\hat{j}) + 0 = 2.0(-1.0\hat{i} + 4.0\hat{j}) + 3.0\mathbf{v}_B (8.0i^+6.0j^)=(−2.0i^+8.0j^)+3.0vB(8.0\hat{i} + 6.0\hat{j}) = (-2.0\hat{i} + 8.0\hat{j}) + 3.0\mathbf{v}_B 3.0vB=10.0i^−2.0j^3.0\mathbf{v}_B = 10.0\hat{i} - 2.0\hat{j} vB=(3.33i^−0.667j^) m s−1\mathbf{v}_B = (3.33\hat{i} - 0.667\hat{j})\,\text{m}\,\text{s}^{-1}

(b) Check kinetic energy:

Before: Ek=12(2.0)(42+32)=1.0×25=25 JE_k = \frac{1}{2}(2.0)(4^2 + 3^2) = 1.0 \times 25 = 25\,\text{J}

After: Ek=12(2.0)(1+16)+12(3.0)(11.11+0.444)=17+12(3.0)(11.56)=17+17.33=34.3 JE_k = \frac{1}{2}(2.0)(1 + 16) + \frac{1}{2}(3.0)(11.11 + 0.444) = 17 + \frac{1}{2}(3.0)(11.56) = 17 + 17.33 = 34.3\,\text{J}

Verifying the kinetic energies:

∣vB∣2=(10/3)2+(−2/3)2=100/9+4/9=104/9=11.56|\mathbf{v}_B|^2 = (10/3)^2 + (-2/3)^2 = 100/9 + 4/9 = 104/9 = 11.56

EkA=12(2.0)(1+16)=17 JE_{kA} = \frac{1}{2}(2.0)(1 + 16) = 17\,\text{J}

EkB=12(3.0)(11.56)=17.33 JE_{kB} = \frac{1}{2}(3.0)(11.56) = 17.33\,\text{J}

Total after =34.33 J= 34.33\,\text{J}, total before =25 J= 25\,\text{J}.

Since 34.33≠2534.33 \ne 25, the collision is not elastic (kinetic energy is not conserved). KE has increased by 9.33 J9.33\,\text{J}, making this a superelastic collision -- energy was released from an internal store (e.g. an explosion).

(c) Impulse on AA =ΔpA=mA(vA−uA)= \Delta\mathbf{p}_A = m_A(\mathbf{v}_A - \mathbf{u}_A)

=2.0[(−1−4)i^+(4−3)j^]=2.0(−5i^+1j^)=(−10i^+2j^) N s= 2.0[(-1 - 4)\hat{i} + (4 - 3)\hat{j}] = 2.0(-5\hat{i} + 1\hat{j}) = (-10\hat{i} + 2\hat{j})\,\text{N}\,\text{s}

Magnitude =100+4=104=10.2 N s= \sqrt{100 + 4} = \sqrt{104} = 10.2\,\text{N}\,\text{s}

UT-2: Coefficient of Restitution and Multiple Collisions​

Question:

A ball of mass 0.15 kg0.15\,\text{kg} is dropped from a height of 2.0 m2.0\,\text{m} onto a horizontal floor. The coefficient of restitution between the ball and the floor is e=0.80e = 0.80.

(a) Calculate the height reached after the first bounce.

(b) Calculate the total time from release until the ball comes to rest. (The ball undergoes infinitely many bounces.)

(c) Calculate the total distance travelled by the ball.

Take g=9.81 m s−2g = 9.81\,\text{m}\,\text{s}^{-2}.

Solution:

(a) Speed just before first impact: v=2gh=◆LB◆2×9.81×2.0◆RB◆=39.24=6.264 m s−1v = \sqrt{2gh} = \sqrt◆LB◆2 \times 9.81 \times 2.0◆RB◆ = \sqrt{39.24} = 6.264\,\text{m}\,\text{s}^{-1}

Speed after first bounce: v′=ev=0.80×6.264=5.012 m s−1v' = ev = 0.80 \times 6.264 = 5.012\,\text{m}\,\text{s}^{-1}

Height after first bounce: h′=v′2/(2g)=25.12/19.62=1.280 mh' = v'^2/(2g) = 25.12/19.62 = 1.280\,\text{m}

Or directly: h′=e2h=0.64×2.0=1.28 mh' = e^2 h = 0.64 \times 2.0 = 1.28\,\text{m}

(b) Time for first descent: t0=2h/g=4.0/9.81=0.6386 st_0 = \sqrt{2h/g} = \sqrt{4.0/9.81} = 0.6386\,\text{s}

Time for first ascent: t1=v′/g=5.012/9.81=0.5109 st_1 = v'/g = 5.012/9.81 = 0.5109\,\text{s}

Since h′=e2hh' = e^2 h and t∝ht \propto \sqrt{h}, each successive bounce time is multiplied by ee.

Total time =t0+2t1+2et1+2e2t1+…= t_0 + 2t_1 + 2et_1 + 2e^2t_1 + \ldots

=t0+2t1(1+e+e2+…)=t0+2t11−e= t_0 + 2t_1(1 + e + e^2 + \ldots) = t_0 + \frac{2t_1}{1 - e}

=0.6386+◆LB◆2×0.5109◆RB◆◆LB◆1−0.80◆RB◆=0.6386+1.02180.20=0.6386+5.109=5.748 s= 0.6386 + \frac◆LB◆2 \times 0.5109◆RB◆◆LB◆1 - 0.80◆RB◆ = 0.6386 + \frac{1.0218}{0.20} = 0.6386 + 5.109 = 5.748\,\text{s}

(c) Total distance =h+2h′+2h′′+…=h+2e2h+2e4h+…= h + 2h' + 2h'' + \ldots = h + 2e^2h + 2e^4h + \ldots

=h+2he2(1+e2+e4+…)=h+2he21−e2= h + 2he^2(1 + e^2 + e^4 + \ldots) = h + \frac{2he^2}{1 - e^2}

=2.0+◆LB◆2×2.0×0.64◆RB◆◆LB◆1−0.64◆RB◆=2.0+2.560.36=2.0+7.11=9.11 m= 2.0 + \frac◆LB◆2 \times 2.0 \times 0.64◆RB◆◆LB◆1 - 0.64◆RB◆ = 2.0 + \frac{2.56}{0.36} = 2.0 + 7.11 = 9.11\,\text{m}

UT-3: Impulse-Momentum with Variable Force​

Question:

A ball of mass 0.40 kg0.40\,\text{kg} is initially at rest. A force F=(6.0t−0.5t2) NF = (6.0t - 0.5t^2)\,\text{N} acts on it for 4.0 s4.0\,\text{s} in a fixed direction.

(a) Calculate the impulse imparted to the ball.

(b) Calculate the velocity of the ball at t=4.0 st = 4.0\,\text{s}.

(c) Calculate the average force over the 4.0 s4.0\,\text{s} interval and explain why it differs from the force at t=2.0 st = 2.0\,\text{s}.

Solution:

(a) Impulse =∫04.0F dt=∫04.0(6.0t−0.5t2) dt= \displaystyle\int_0^{4.0} F\,dt = \int_0^{4.0} (6.0t - 0.5t^2)\,dt

=[3.0t2−t36]04=(3.0×16−646)−0=(48−10.67)=37.33 N s= \left[3.0t^2 - \frac{t^3}{6}\right]_0^4 = (3.0 \times 16 - \frac{64}{6}) - 0 = (48 - 10.67) = 37.33\,\text{N}\,\text{s}

(b) By the impulse-momentum theorem: J=Δp=mv−0J = \Delta p = mv - 0

v=J/m=37.33/0.40=93.3 m s−1v = J/m = 37.33/0.40 = 93.3\,\text{m}\,\text{s}^{-1}

(c) Average force =J/Δt=37.33/4.0=9.33 N= J/\Delta t = 37.33/4.0 = 9.33\,\text{N}

Force at t=2.0 st = 2.0\,\text{s}: F(2)=6.0(2)−0.5(4)=12−2=10 NF(2) = 6.0(2) - 0.5(4) = 12 - 2 = 10\,\text{N}

The average force (9.33 N9.33\,\text{N}) differs from the force at the midpoint (10 N10\,\text{N}) because the force is not linear in time -- it is quadratic. For a linearly varying force, the average equals the midpoint value, but for a quadratic force this is not the case. The average force is determined by the integral over the entire interval, weighted by time.

Integration Tests​

IT-1: Oblique Collision with a Wall (with Kinematics)​

Question:

A ball of mass 0.20 kg0.20\,\text{kg} hits a smooth vertical wall with velocity (8.0i^−6.0j^) m s−1(8.0\hat{i} - 6.0\hat{j})\,\text{m}\,\text{s}^{-1}, where i^\hat{i} is horizontal (perpendicular to the wall) and j^\hat{j} is vertical (parallel to the wall). The coefficient of restitution between the ball and wall is e=0.75e = 0.75. The wall is smooth, so there is no friction.

(a) Calculate the velocity of the ball after the collision.

(b) Calculate the impulse exerted by the wall on the ball.

(c) After rebounding, the ball moves freely under gravity. Calculate the horizontal distance from the wall at which the ball returns to its original height.

Take g=9.81 m s−2g = 9.81\,\text{m}\,\text{s}^{-2}.

Solution:

(a) The wall is vertical and smooth. The j^\hat{j}-component (parallel to the wall) is unchanged. The i^\hat{i}-component (perpendicular to the wall) reverses with restitution:

vx′=−e⋅ux=−0.75×8.0=−6.0 m s−1v_x' = -e \cdot u_x = -0.75 \times 8.0 = -6.0\,\text{m}\,\text{s}^{-1}

vy′=uy=−6.0 m s−1v_y' = u_y = -6.0\,\text{m}\,\text{s}^{-1}

Velocity after collision: (−6.0i^−6.0j^) m s−1(-6.0\hat{i} - 6.0\hat{j})\,\text{m}\,\text{s}^{-1}

(b) Impulse =m(v′−u)= m(\mathbf{v}' - \mathbf{u})

=0.20[(−6.0−8.0)i^+(−6.0−(−6.0))j^]= 0.20[(-6.0 - 8.0)\hat{i} + (-6.0 - (-6.0))\hat{j}] =0.20(−14.0i^+0j^)=(−2.8i^) N s= 0.20(-14.0\hat{i} + 0\hat{j}) = (-2.8\hat{i})\,\text{N}\,\text{s}

The impulse is purely horizontal (perpendicular to the wall), as expected for a smooth wall.

(c) The ball rebounds with vx=−6.0 m s−1v_x = -6.0\,\text{m}\,\text{s}^{-1} (away from wall) and vy=−6.0 m s−1v_y = -6.0\,\text{m}\,\text{s}^{-1} (downward).

Time to return to original height: Δy=0=vyt+12(−g)t2\Delta y = 0 = v_y t + \frac{1}{2}(-g)t^2

0=−6.0t−4.905t2⇒t(−6.0−4.905t)=00 = -6.0t - 4.905t^2 \Rightarrow t(-6.0 - 4.905t) = 0

Ignoring t=0t = 0: t=−6.0/4.905=−1.223 st = -6.0/4.905 = -1.223\,\text{s}

This is negative, which means the ball is below its original height and moving downward. The ball never returns to its original height -- it hits the ground first.

If the collision happens at height hh above the ground, we need hh. Since no height is specified, we note that with vy=−6.0 m s−1v_y = -6.0\,\text{m}\,\text{s}^{-1} downward and decelerating (gravity acts downward), the ball continues to descend. The ball would need an upward component to return to its original height. Since vyv_y is negative, the ball never rises back to its collision height.

Horizontal distance from wall at time tt: x=∣vx∣×t=6.0tx = |v_x| \times t = 6.0t. But since the ball never returns to its original height, the question's premise requires that the collision occurred at ground level, or the ball was given height. If we interpret "returns to its original height" as asking about the trajectory, the ball cannot return since it is moving downward with no upward component.

Answer: The ball cannot return to its original height after the collision since its vertical velocity component is directed downward and gravity does not reverse it.

IT-2: Rocket Propulsion and Momentum (with Dynamics)​

Question:

A rocket of mass 500 kg500\,\text{kg} (including fuel) is initially at rest in deep space (no external forces). It burns fuel at a rate of 5.0 kg s−15.0\,\text{kg}\,\text{s}^{-1}, ejecting exhaust gas at a speed of 2000 m s−12000\,\text{m}\,\text{s}^{-1} relative to the rocket.

(a) Calculate the thrust produced by the rocket engine.

(b) Calculate the acceleration of the rocket at t=0t = 0 and at t=60 st = 60\,\text{s}.

(c) Calculate the speed of the rocket after 60 s60\,\text{s} of burning.

Solution:

(a) Thrust =dmdt×vexhaust=5.0×2000=10000 N= \frac{dm}{dt} \times v_{\text{exhaust}} = 5.0 \times 2000 = 10000\,\text{N}

(b) At t=0t = 0: mass =500 kg= 500\,\text{kg}

a0=◆LB◆Fthrust◆RB◆◆LB◆m◆RB◆=10000500=20.0 m s−2a_0 = \frac◆LB◆F_{\text{thrust}}◆RB◆◆LB◆m◆RB◆ = \frac{10000}{500} = 20.0\,\text{m}\,\text{s}^{-2}

At t=60 st = 60\,\text{s}: fuel burned =5.0×60=300 kg= 5.0 \times 60 = 300\,\text{kg}, remaining mass =200 kg= 200\,\text{kg}

a60=10000200=50.0 m s−2a_{60} = \frac{10000}{200} = 50.0\,\text{m}\,\text{s}^{-2}

The acceleration increases with time because the rocket's mass decreases while the thrust remains constant.

(c) Using the rocket equation (Tsiolkovsky):

Δv=vexhaustln⁡(m0mf)=2000ln⁡(500200)=2000ln⁡(2.5)=2000×0.9163=1833 m s−1\Delta v = v_{\text{exhaust}} \ln\left(\frac{m_0}{m_f}\right) = 2000 \ln\left(\frac{500}{200}\right) = 2000 \ln(2.5) = 2000 \times 0.9163 = 1833\,\text{m}\,\text{s}^{-1}

Alternatively, using impulse: the total momentum of all exhaust gas ejected provides the momentum gain of the rocket. The difficulty is that the exhaust speed relative to an inertial frame changes as the rocket accelerates, so the integral form (rocket equation) is needed.

Check using approximation (treating exhaust velocity as constant in lab frame, which gives a lower bound):

Total fuel ejected =300 kg= 300\,\text{kg}. Total impulse =300×2000=600000 N s= 300 \times 2000 = 600000\,\text{N}\,\text{s}.

If all went to the rocket: Δv=600000/500=1200 m s−1\Delta v = 600000/500 = 1200\,\text{m}\,\text{s}^{-1} (underestimate since it ignores the rocket's decreasing mass).

The correct answer from the rocket equation is 1833 m s−11833\,\text{m}\,\text{s}^{-1}.

IT-3: Collision Analysis with Energy Dissipation (with Work-Energy)​

Question:

A bullet of mass 10 g10\,\text{g} is fired horizontally with speed 400 m s−1400\,\text{m}\,\text{s}^{-1} into a wooden block of mass 1.99 kg1.99\,\text{kg} resting on a rough horizontal surface. The bullet embeds itself in the block. The coefficient of friction between the block and the surface is μ=0.30\mu = 0.30.

(a) Calculate the speed of the block (with embedded bullet) immediately after the collision.

(b) Calculate the distance the block slides before coming to rest.

(c) Calculate the energy lost in the collision and the energy lost due to friction. Express each as a percentage of the bullet's initial kinetic energy.

Take g=9.81 m s−2g = 9.81\,\text{m}\,\text{s}^{-2}.

Solution:

(a) Conservation of momentum (perfectly inelastic collision):

mbvb=(mb+mB)Vm_b v_b = (m_b + m_B)V 0.010×400=(0.010+1.99)V0.010 \times 400 = (0.010 + 1.99)V 4.0=2.00V4.0 = 2.00V V=2.00 m s−1V = 2.00\,\text{m}\,\text{s}^{-1}

(b) Using the work-energy principle for the sliding block:

12(2.00)V2=μ(2.00)g×d\frac{1}{2}(2.00)V^2 = \mu(2.00)g \times d 12(2.00)(4.0)=0.30×2.00×9.81×d\frac{1}{2}(2.00)(4.0) = 0.30 \times 2.00 \times 9.81 \times d 4.0=5.886d4.0 = 5.886d d=0.680 md = 0.680\,\text{m}

(c) Initial KE of bullet: Ek0=12(0.010)(400)2=800 JE_{k0} = \frac{1}{2}(0.010)(400)^2 = 800\,\text{J}

KE after collision: Ek1=12(2.00)(2.00)2=4.0 JE_{k1} = \frac{1}{2}(2.00)(2.00)^2 = 4.0\,\text{J}

Energy lost in collision: 800−4.0=796 J800 - 4.0 = 796\,\text{J}

Percentage: 796800×100=99.5%\frac{796}{800} \times 100 = 99.5\%

Energy lost due to friction: =Ek1=4.0 J= E_{k1} = 4.0\,\text{J} (all remaining KE is lost to friction)

Percentage: 4.0800×100=0.50%\frac{4.0}{800} \times 100 = 0.50\%

This demonstrates why perfectly inelastic collisions are so inefficient: 99.5%99.5\% of the bullet's kinetic energy is converted to heat, sound, and deformation in the collision.