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What is Coefficient of Restitution With its Formula?
In an inelastic collision, the total kinetic energy is not con served, but the total momentum is conserved.
Two particles are moving in the same direction in a straight line with velocities u1 and u2 respectively (u1 > u2) [Fig.]. Before the collision, the distance between the two particles will decrease, and their velocity of approach will be = (u1 – u2). After collision, their separation increases with time i.e., (v2 > v1) where v1 = velocity of first particle, v2 = velocity of second particle after the impact. Therefore, (v2 – v1) = the velocity of separation.
Coefficient of restitution: Coefficient of restitution is defined as the ratio of the velocity with which the two bodies separate after collision to the velocity of their approach before collision.
i.e., e \(=\frac{\text { relative velocity of separation after collision }}{\text { relative velocity of approach before collision }}\)
= \(\frac{v_2-v_1}{u_1-u_2}\)
Perfectly inelastic collision: In this type of collision the two colliding bodies stick together and move as a single body, i.e., they move with a common velocity. In the Fig., a perfectly inelastic collision has been shown.
Suppose, two bodies of masses m1 and m2, moving with velocities u1 and u2 respectively in the same direction along a straight line, collide perfectly inelastically. After collision they stick together and move with velocity v along the same straight line.
From the principle of conservation of linear momentum we get,
m1u1 + m2u2 = (m1 + m2)v
or, v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\) …… (2)
Loss of kinetic energy:
Total kinetic energy before collision = \(\frac{1}{2} m_1 u_1^2\) + \(\frac{1}{2} m_2 u_2^2\)
Total kinetic energy after collision = \(\frac{1}{2}\)(m1 + m2)v2
∴ Loss of kinetic energy
Here (u1 – u2) is the relative velocity of the two bodies before collision. So the loss of kinetic energy is proportional to the square of this relative velocity Again if u2 = 0 i.e., if the second body is at rest before collision the loss of kinetic energy becomes maximum.
Partially elastic collision: In this type of collision the two bodies do not stick, but move separately along the same straight line with different velocities [Fig.]. Here momentum remains conserved while the total kinetic energy decreases.
The relative velocity of the two bodies before collision = u1 – u2 and that after collision = v2 – v1.
So the coefficient of restitution,
e = \(\frac{v_2-v_1}{u_1-u_2}\) …… (4)
According to the law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
Now from equation (4) we have, e(u1 – u2) = v2 – v1
Multiplying the above equation by m2 we have,
em2u1 – em2u2 = m2v2 – m2v1 …… (6)
Subtracting equation (6) from equation (5) we get,
u1(m1 – em2) + u2(1 + e)m2 = (m1 + m2)v1
∴ v1 = \(\frac{\left(m_1-e m_2\right) u_1+(1+e) m_2 u_2}{m_1+m_2}\) ………. (7)
Similarly, v2 = \(\frac{\left(m_2-e m_1\right) u_2+(1+e) m_1 u_1}{m_1+m_2}\) …….. (8)
Loss of kinetic energy:
- If e = 1 , i.e., if the collision is perfectly elastic, then the loss of kinetic energy = 0, i.e., in case of a perfectly elastic collision kinetic energy remains conserved.
- If e = 0, i.e., if the collision is perfectly inelastic, kinetic energy decreases and the loss of kinetic energy
= \(\frac{m_1 m_2}{2\left(m_1+m_2\right)}\)(u1 – u2)2 - For partially elastic collision, 0 < e < 1.
Numerical Examples
Example 1.
Two bodies of masses 5 kg and 10 kg move towards each other with velocities 10 m ᐧ s-1 and 14 m ᐧ s-1 respectively. If the coefficient of restitution is 0.8, find their velocities after the collision.
Solution:
Let u1 and u2 respectively be the velocities of the first and the second body before the collision. After the collision, their velocities will be v1 and v2 respectively.
From the law of conservation of linear momentum
m1u1 + m2u2 = m1v1 + m2v2
Here, m1 = 5kg, m2 = 10kg, u1 = 10m ᐧ s-1 and u2 = -14m ᐧ s-1
[let us assume u1 to be taken along positive x -axis and u2 along negative x-axis]
∴ 5 × 10 + 10 × (-14) = 5v1 + 10v2
or, v1 + 2v2 = -18
Also, e = \(\frac{v_2-v_1}{u_1-u_2}\) or, 0.8 = \(\frac{v_2-v_1}{10-(-14)}\)
or, v2 – v1 = 0.8 × 24 = 19.2
Hence, from (1) and (2),
3v2 = 1.2 or, v2 = \(\frac{1.2}{3}\) = 0.4 m ᐧ s-1
and v1 = -18.8 m ᐧ s-1 which means after the collision the 5 kg mass advances along negative x -axis and the 10 kg mass goes along positive x -axis.
Example 2.
Two bodies of masses 1 kg and 0.5 kg move towards each other with velocities 10 cm ᐧ s-1 and 5 cm ᐧ s-1 respectively. After collision, the bodies coalesce (join together). Find the common velocity after collision and the loss in kinetic energy.
Solution:
Let the velocity of the combined mass after collision be v. From the law of conservation of momentum,
m1u1 + m2u2 = (m1 + m2)v
or, 1 × 0.1 + 0.5 × (-0.05) = (1 + 0.5)v
or, 0.1 – 0.025 = 1.5v or, v = \(\frac{0.075}{1.5}\) = 0.05m ᐧ s-1
This implies that the velocity of the combined mass is along the direction of the initial velocity of the 1 kg mass.
Initial kinetic energy of the bodies
= \(\frac{m_1 u_1^2}{2}+\frac{m_2 u_2^2}{2}\) = \(\frac{1}{2}\) × 1 × (0.1)2 × \(\frac{1}{2}\) × (0.5) × (0.05)2
= 5.625 × 10-3 J
Final kinetic energy of the combined mass after collision
= \(\left(m_1+m_2\right) \frac{v^2}{2}\)
= \(\frac{1}{2}\) × (1 + 0.5) × (0.05)2 = 1.875 × 10-3 J
∴ Loss in kinetic energy
= (5.625 – 1.875) × 10-3 = 3.75 × 10-3 J.
Example 3.
A car of mass 2000 kg collides with a truck of mass 104 kg moving at 48 km ᐧ h-1. After collision, the car rides up the truck and the truck-car combina-tion moves at 15 km ᐧ h-1. What was the velocity of the car before collision?
Solution:
The collision is inelastic as both the bodies move together after collision. Suppose a body of mass m, moving with velocity u, collides inelastically along a straight line with a body of mass M and velocity v. After collision, the two masses combine and move with velocity V. Applying the law of conservation of momentum,
mu + Mv = (m + M)V or, mu = (m + M)V – Mv
or, u = \(\frac{m+M}{m} V\) – \(\frac{M}{m} v\) = \(\left(1+\frac{M}{m}\right) V\) – \(\frac{M}{m} v\)
Here m = 2000kg, v = 48km ᐧ h-1, M = 10000 kg, V = 15km ᐧ h-1
∴ \(\frac{M}{m}\) = \(\frac{10000}{2000}\) = 5
u = (1 + 5) × 15 – 5 × 48 = 90 – 240 = -150km ᐧ h-1
The negative sign indicates that, before collision, the car was moving in the direction opposite to that of the truck.
Example 4.
A ball of mass 100 g was thrown vertically upwards with a velocity of 49 m ᐧ s-1. At the same time another identical ball was dropped, to fall along the same path from a height of 98 m. After some time, the two balls collided and got stuck together. This combined mass then fell to the ground. How long was the combined mass in motion?
Solution:
Suppose the balls collide at a height h above the ground in time t1. Analysing the upward motion of the 1st ball, we get,
h = 49 t1 – 9.8 \(t_1^2 / 2\) ……. (1)
Analysing the downward motion of the 2nd ball, we get,
98 – h = 9.8 \(t_1^2 / 2\) …… (2)
From equations (1) and (2),
98 = 49t1 or, t1 = 2 s
At the time of collision, let the velocity of the 1st ball be v1 and that of the 2nd ball be v2.
∴ v1 = 49 – 9.8 × 2 = 29.4 m ᐧ s-1 (upward)
and v2 = 9.8 × 2 = 19.6 m ᐧ s-1 (downward)
After collision, let the velocity of the combined mass be V.
Then according to the law of conservation of momentum,
0.1 × 29.4 – 0.1 × 19.6 = 2 × 0.1 × V
[indicating downward motion with a negative sign]
or, V = 4.9 m ᐧ s-1 (upward)
From equation (1), we get,
h = 49 × 2 – 9.8 × (2)2/2 = 78.4 m
If the combined mass was in motion for a time t, then from the relation h = ut + \(\frac{1}{2}\)gt2, we get,
-78.4 = 4.9t – \(\frac{1}{2}\) × 9.8t2, or, t2 – t – 16 = 0
∴ t = \(\frac{1}{2}(1 \pm \sqrt{1+4 \times 1 \times 16})\)
As t cannot be negative, we have, t = 4.53 s
Example 5.
A bullet of mass 50 g is fired into a wooden block of mass 2 kg resting on a smooth table surface. The
bullet enters at 50 m ᐧ s-1 and gets embedded in the block. Find the final velocity of the block. Find the initial and the final kinetic energy of the block-bullet system. [HS(XI) ‘06]
Solution:
Look at Fig. From the law of conservation of momentum,
Here, m1 = 50 g = 0.05 kg, v = 50 m ᐧ s-1 and m2 = 2 kg.
∴ The final velocity of the block
V = \(\frac{0.05 \times 50}{0.05+2}\) = 1.22 m ᐧ s-1
Initial kinetic energy of the system
Ei = \(\frac{1}{2}\) × 0.05 × 502 = 62.5J.
Final kinetic energy,
Ef = \(\frac{1}{2}\) × (2 + 0.05) × (1.22)2 = 1.52 J
Example 6.
A bullet of mass 250 g, moving with a horizontal velocity of 400 m ᐧ s-1, gets embedded in a target. The target of mass 4.75 kg can move freely. Find the loss of kinetic energy due to this collision. What happens to this energy loss?
Solution:
Mass of the bullet, m = 250 g = 0.25 kg,
velocity of the bullet, u = 400 m ᐧ s-1, and mass of the target, M = 4.75 kg.
After the impact, the target and the bullet both move with a velocity V (say).
mv = (M + m)V
∴ V = \(\frac{m v}{M+m}\) = \(\frac{0.25 \times 400}{0.25+4.75}\) = 20 m ᐧ s-1.
Kinetic energy of the target and the bullet after collision
= \(\frac{1}{2}\)(M + m)V2 = \(\frac{1}{2}\)(4.75 + 0.25) × 202 = 1000 J
Kinetic energy of the bullet before collision,
\(\frac{1}{2}\)mv2 = \(\frac{1}{2}\) × 0.25 × (400)2 = 20000 J.
Hence, the kinetic energy lost in collision
= 20000 – 1000 = 19000 J.
This kinetic energy transforms into heat and sound energy.