Worked examples · Sheet 3

Perfectly inelastic collisions and ballistic pendulums

Theory used: Collision Perfectly inelastic collision Ballistic pendulum Conservation of momentum Conservation of a single component König's theorem for kinetic energy Energy that can be dissipated in a collision

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

On a straight horizontal track with negligible friction, a cart $A$ of mass $m_{A} = 2.0 kg$ moves at $3.0 m/s$ to the right and a cart $B$ of mass $m_{B} = 1.0 kg$ moves at $3.0 m/s$ to the left. The carts collide and stay coupled. Compute the velocity after the collision and the dissipated kinetic energy.

x axis to the right: v_{A} = 3.0 m/s, v_{B} = - 3.0 m/s

m_{A} v_{A} + m_{B} v_{B} = (m_{A} + m_{B}) V ⟹ V = \frac{m _{A} v _{A} + m _{B} v _{B}}{m _{A} + m _{B}}

V = \frac{2.0 \cdot 3.0 + 1.0 \cdot ( - 3.0 )}{3.0} m/s = \frac{3.0}{3.0} m/s = 1.0 m/s

K_{i} = \frac{1}{2} \cdot 2.0 \cdot 3. 0^{2} J + \frac{1}{2} \cdot 1.0 \cdot 3. 0^{2} J = 9.0 J + 4.5 J = 13.5 J

K_{f} = \frac{1}{2} (m_{A} + m_{B}) V^{2} = \frac{1}{2} \cdot 3.0 \cdot 1. 0^{2} J = 1.5 J, K_{i} - K_{f} = 12 J

The track is horizontal and smooth, so the external forces on the two carts, that is the weights and the reactions of the track, are vertical, and along $x$ the resultant of the external forces is zero. Therefore the $x$ component of momentum is conserved, even though during the collision the couplings deform and dissipate energy. The carts stay coupled, so the collision is perfectly inelastic and after the collision there is a single unknown, the common velocity $V$ .

The velocities have opposite signs, therefore the momenta partly cancel and the final velocity is $1.0 m/s$ to the right, the direction of the cart with the larger momentum in magnitude. The kinetic energy goes from $13.5 J$ to $1.5 J$ , so the collision dissipates $12 J$ , almost $90 %$ of the initial energy.

As a check, the formula $\frac{1}{2} \frac{m _{A} m _{B}}{m _{A} + m _{B}} (v_{A} - v_{B})^{2} = \frac{1}{2} \cdot \frac{2}{3} \cdot 36 J = 12 J$ gives the same value.

Exercise 2

A bullet of mass $m = 8.0 g$ travels horizontally and embeds itself in a wooden block of mass $M = 1.6 kg$ , at rest and hanging from two vertical inextensible strings of negligible mass. After the collision the block rises to a height $h = 12 cm$ above the initial position. Neglecting air resistance, compute the speed of the bullet and the fraction of the initial kinetic energy dissipated in the collision.

Diagram of the ballistic pendulum (not to scale): perfectly inelastic collision at the lowest position, then rise of the system up to height

h

⎩ ⎨ ⎧ m v = (m + M) V \frac{1}{2} (m + M) V^{2} = (m + M) g h ⟹ v = \frac{m + M}{m} 2 g h

V = 2 \cdot 9.81 \cdot 0.12 m/s \approx 1.53 m/s

v = \frac{1.608 kg}{0.0080 kg} \cdot 1.53 m/s \approx 3.1 \cdot 1 0^{2} m/s

\frac{K _{i} - K _{f}}{K _{i}} = 1 - \frac{\frac{1}{2} ( m + M ) V ^{2}}{\frac{1}{2} m v ^{2}} = 1 - \frac{m}{m + M} = \frac{M}{m + M} = \frac{1.6}{1.608} \approx 0.995

The problem contains two phases, governed by different laws. In the collision the strings are vertical, so weights and tensions have no horizontal component and the horizontal momentum is conserved; the bullet stays in the block, so the collision is perfectly inelastic and mechanical energy in this phase is not conserved. After the collision the tension of the strings is perpendicular to the velocity and does no work, and air is negligible; therefore the only force that does work is the weight and the mechanical energy of the block with the bullet is conserved.

The second equation of the system gives the velocity $V$ immediately after the collision, which does not depend on the masses; substituted into the first, it gives the speed of the bullet, about $3.1 \cdot 1 0^{2} m/s$ .

The ratio between the kinetic energies after and before the collision, using $V = \frac{m v}{m + M}$ , equals $\frac{m}{m + M}$ . Therefore in the collision $99.5 %$ of the kinetic energy of the bullet is dissipated, becoming deformation and heat in the wood. Applying conservation of energy between the bullet in flight and the highest position would lead to a speed of about $1.53 m/s \cdot \frac{1.608}{0.0080} \approx 22 m/s$ , a value wrong by more than an order of magnitude.

Exercise 3

A bullet of mass $m = 20 g$ , with horizontal velocity $v = 250 m/s$ , embeds itself in a block of mass $M = 2.48 kg$ at rest on a horizontal plane. The coefficient of kinetic friction between block and plane is $μ_{d} = 0.40$ . Compute the velocity of the block immediately after the collision and the distance it travels before stopping.

m v = (m + M) V ⟹ V = \frac{m v}{m + M}

- μ_{d} (m + M) g d = 0 - \frac{1}{2} (m + M) V^{2} ⟹ d = \frac{V ^{2}}{2 μ _{d} g}

V = \frac{0.020 \cdot 250}{2.50} m/s = 2.0 m/s

d = \frac{( 2.0 m/s ) ^{2}}{2 \cdot 0.40 \cdot 9.81 m/ s ^{2}} \approx 0.51 m

During the collision friction also acts on the block, and it is a horizontal external force. Friction, however, has bounded magnitude, at most $μ_{d} (m + M) g \approx 10 N$ , and it is a non-impulsive force; in the very short interval of the collision its impulse is negligible compared with the momentum of the bullet, $5.0 kg \cdot m/s$ . Therefore in the collision the horizontal momentum is conserved and the block starts with $V = 2.0 m/s$ .

After the collision the block with the bullet slides on the plane and kinetic friction does negative work, equal to $- μ_{d} (m + M) g d$ , while weight and normal reaction do no work because they are perpendicular to the displacement. By the work-energy theorem this work equals the change in kinetic energy, which goes from $\frac{1}{2} (m + M) V^{2}$ to zero. The mass cancels and the block stops after about $0.51 m$ .

As a dimensional check, $\frac{V ^{2}}{g}$ has the dimensions of $\frac{m ^{2} / s ^{2}}{m/ s ^{2}} = m$ , and $μ_{d}$ is a pure number.

Exercise 4

Two bodies of masses $m_{A} = 3.0 kg$ and $m_{B} = 2.0 kg$ move along the same horizontal line, on a smooth plane: $A$ with velocity $5.0 m/s$ , $B$ at rest. Using König’s theorem, compute the kinetic energy of the motion of the center of mass and the kinetic energy in the center-of-mass frame. Deduce the dissipated energy if the two bodies collide and stick together.

v_{c m} = \frac{m _{A} v _{A} + m _{B} v _{B}}{m _{A} + m _{B}} = \frac{3.0 \cdot 5.0}{5.0} m/s = 3.0 m/s

K = \frac{1}{2} \cdot 3.0 \cdot 5. 0^{2} J = 37.5 J, \frac{1}{2} M v_{c m}^{2} = \frac{1}{2} \cdot 5.0 \cdot 3. 0^{2} J = 22.5 J

K^{*} = K - \frac{1}{2} M v_{c m}^{2} = 37.5 J - 22.5 J = 15 J

v_{A}^{*} = 5.0 - 3.0 = 2.0 m/s, v_{B}^{*} = 0 - 3.0 = - 3.0 m/s

K^{*} = \frac{1}{2} \cdot 3.0 \cdot 2. 0^{2} J + \frac{1}{2} \cdot 2.0 \cdot 3. 0^{2} J = 6.0 J + 9.0 J = 15 J

By König’s theorem the kinetic energy of the system is the sum of the energy of the overall motion, $\frac{1}{2} M v_{c m}^{2}$ , and of the energy $K^{*}$ computed in the center-of-mass frame. We first compute the velocity of the center of mass, $3.0 m/s$ , then the overall part, $22.5 J$ , and by difference $K^{*} = 15 J$ . The direct calculation with the velocities relative to the center of mass, $v_{i}^{*} = v_{i} - v_{c m}$ , confirms the value; the two momenta in the center-of-mass frame, $6.0$ and $- 6.0 kg \cdot m/s$ , have zero sum, as they must.

The plane is smooth, so in the collision no horizontal external forces act and $v_{c m}$ remains $3.0 m/s$ ; consequently the part $\frac{1}{2} M v_{c m}^{2} = 22.5 J$ cannot be dissipated. If the bodies stick together, after the collision they are at rest relative to the center of mass and $K^{*}$ vanishes. Therefore the dissipated energy is $K^{*} = 15 J$ , the maximum that a collision between these two bodies can dissipate.