Physics Problem

Mechanics · symbolic answer, dimension check, limiting case, then numbers

1. Projectile range

A projectile leaves the ground at speed (v_0) and angle θ.

Horizontal and vertical motion decouple. Time of flight from y(t)=(v_0)*sin(θ)*t−1/2*g*t^2=0:

(t_ƒ)=(2*(v_0)*sin(θ))/g

R=(v_0)*cos(θ)⋅(t_ƒ)=(2*(v_0)^2*sin(θ)*cos(θ))/g=((v_0)^2*sin(2)*θ)/g

Dimensions. [(v_0)^2/g]=(m^2*s^(−2))/(m*s^(−2))=m

Limits. R=0 at θ=0 and θ=90° maximum at θ=45° where sin(2)*θ=1. Both correct.

Numbers. (v_0)=25m/s θ=35°

R=(625*sin(70°))/9.81=(625*(0.9397))/9.81≈59.9m

2. Block on an incline with friction

Mass m slides down a slope at angle θ kinetic friction coefficient (μ_k).

Take axes along and normal to the slope, which removes the need to decompose the normal force at all.

Normal: N=m*g*cos(θ)

Along: m*a=m*g*sin(θ)−(μ_k)*N=m*g*sin(θ)−(μ_k)*m*g*cos(θ)

a=g*(sin(θ)−(μ_k)*cos(θ))

The mass cancels, as it must — friction and weight are both proportional to m.

Limits. a=0 when (μ_k)=tan(θ) the angle of repose. Above that the expression goes negative, which does not mean the block accelerates uphill: it means the block was never sliding and kinetic friction was the wrong model.

Numbers. θ=30°, (μ_k)=0.25

a=9.81*(0.5−0.25*(0.8660))=9.81*(0.2835)≈2.78m/s^2

3. Pendulum at the lowest point

Length L released from rest at angle (θ_0).

Tension does no work, being always perpendicular to the motion, and there is no friction, so mechanical energy is conserved. Height dropped:

h=L−L*cos((θ_0))=L*(1−cos((θ_0)))

1/2*m*v^2=m*g*h⟹v=√(,2*g*L*(1−cos((θ_0))))

Dimensions. √(,m*s^(−2)⋅m)=m/s

Limits. (θ_0)=0 gives v=0. (θ_0)=90° gives v=√(,2*g*L) the free-fall speed after dropping L which is right since the bob has fallen exactly L.

Numbers. L=1.5m (θ_0)=40°

v=√(,2*(9.81)*(1.5)*(1−0.766))=√(,6.888)≈2.62m/s

Note this is exact for any (θ_0). The small-angle approximation is needed for the period, not for this.

4. Car on an unbanked curve

Mass m radius r static friction coefficient (μ_s).

Friction supplies the entire centripetal force, and it can supply at most (μ_s)*N=(μ_s)*m*g:

(m*v^2)/r≤(μ_s)*m*g⟹(v_max)=√(,(μ_s)*g*r)

Mass cancels again. A loaded truck and an empty one slip at the same speed, which is counterintuitive and worth remembering.

Numbers. r=50*m (μ_s)=0.7

(v_max)=√(,0.7*(9.81)*(50))=√(,343.35)≈18.5*m/s≈66.7km/h

5. Atwood machine

Masses (m_1)>(m_2) over a frictionless massless pulley.

(m_1)*g−T=(m_1)*a,T−(m_2)*g=(m_2)*a

Adding eliminates T

a=(((m_1)−(m_2))*g)/((m_1)+(m_2)),T=(2*(m_1)*(m_2)*g)/((m_1)+(m_2))

Limits. (m_1)=(m_2) gives a=0 and T=m*g a balanced system. (m_2)=0 gives a=g and T=0 free fall on a slack string. (m_2)→∞ gives a→−g.

T is the harmonic mean of the weights up to a factor, so it always sits between (m_2)*g and (m_1)*g — the string cannot pull harder than the heavier weight or slacker than the lighter one.