Studying Physics Topics can lead to exciting new discoveries and technological advancements.
What is the Velocity Attained When a Ball is Dropped from Height h?
Let a sphere be dropped from a height h onto a horizontal plane at rest [Fig.]. After the impact with the plane, the sphere bounces up to a height H, (H < h). It falls again and after a second impact with the plane the sphere comes to rest on the plane. These collisions are inelastic.
As the sphere is dropped from a height h, the velocity u with which it strikes the plane, H
v = \(\sqrt{2 g h}\) …….. (1)
If the coefficient of restitution for the sphere-plane impact is e, the velocity of separation = ev. Hence, after the first collision, the sphere rises up with the velocity ev. If it rises up to the height H,
0 = (ev)2 – 2gH
or, H = \(\frac{e^2 \cdot 2 g h}{2 g}\) = e2h ……. (2)
If the collision was elastic, e = 1 and hence, H = h, i.e., the sphere would again rise up to h.
Total distance travelled before coming to rest: After the first impact, the sphere rises up to a height H, given by (2). Then it fails from that height and causes a second impact with the plane. After the second impact, if the sphere rises up to a height H’ then from (2), the heights attained after the 2nd, 3rd, … impacts are respectively,
H’ = e2H = e2ᐧe2h = e4h,
H”= e2H’ = e2ᐧe4h, and so on.
Hence, considering the up and down motions, the distance travelled, d, before coming to rest is given by the series,
d = h + 2e2h + 2e4h + 2e6h + …….. ∞
= h + 2e2h(1 + e2 + e4 + e6 + …..∞)
= h + 2e2h\(\frac{1}{1-e^2}\)
[since the collisions are inelastic, e2 < 1 here.]
= h\(\left[1+\frac{2 e^2}{1-e^2}\right]\)
= h\(\frac{1+e^2}{1-e^2}\) ……. (3)
Time taken to reach the steady state: Let t0 denote the time taken by the sphere to come down from the height h.
∴ h = \(\frac{g t_0^2}{2}\) or, t0 = \(\sqrt{\frac{2 h}{g}}\)
If the time interval between the first and the second impacts is T, then,
0 = evᐧT – \(\frac{g T^2}{2}\)
or, gT2 = 2evT
or, T = \(\frac{2 e v}{g}\) = \(\frac{2 e \sqrt{2 g h}}{g}\) = 2e\(\sqrt{\frac{2 h}{g}}\)
Similarly, if T’ is the time interval between the second and the third impacts then
T’ = 2e2\(\sqrt{\frac{2 h}{g}}\)
Hence, total time t taken by the sphere to reach the steady state,
The value of the coefficient of restitution: From the relation H = e2h, e can be calculated by measuring H and h. However, if the same rubber ball is dropped from the same height h, it might not bounce up to H every time. Hence, the value of e determined by this method is an approximate one.
Impulse of the force applied by the plane: Let the mass of the sphere be m, downward velocity (while falling from a height h) just before the first impact be u, and upward velocity just after the impact be ev.
Hence, impulse of the upward force applied by the plane
= change in momentum
= m[eu – (-v)] = mu(1 + e) = m\(\sqrt{2 g h}\)[1 + e]
The loss of kinetic energy: The loss of kinetic energy due to the first impact
= \(\frac{1}{2} m v^2\) – \(\frac{1}{2} m(e v)^2\)
= \(\frac{1}{2} m v^2\)(1 – e2) = \(\frac{1}{2}\)m ᐧ 2gh(1 – e2)
= mgh(1 – e2)
Numerical Examples
Example 1.
A glass ball is dropped from a height of 15 m onto a horizontal glass plate at rest. Find the upward rise of the ball after impact. The coefficient of restitution e = 0.8.
Solution:
If the ball falls from a height h onto a horizontal plane, and rises up to H after impact, then
H = e2h = (0.8)2 × 15 = 9.6 m.
Example 2.
A ball is dropped from a height of 90 m on to a horizontal plane at rest. Find the total distance travelled by it before coming to rest. The coefficient of restitution, e = 0.5.
Solution:
Total distance travelled by the ball
= h × \(\frac{1+e^2}{1-e^2}\) = 90 × \(\frac{1+(0.5)^2}{1-(0.5)^2}\)
= 90 × \(\frac{1.25}{0.75}\) = 150m.
Example 3.
A ball is dropped from a height of 1 m onto a horizontal plane. The ball takes 1.3 s from its time of release for the second impact with the plane. Find the coefficient of restitution.
Solution:
Let the velocity of the ball be V just before the 1st impact with the plane.
Then V2 = 2gh = 2 × 980 × 100
or, V = \(\sqrt{2 \times 980 \times 100}\) = 442.7 cm ᐧ s-1
If the time taken by the ball is t, to fall from the height of 1 m, then
t = \(\sqrt{\frac{2 h}{g}}\) [using h = \(\frac{1}{2} g t^2\)]
= \(\sqrt{2 \times \frac{100}{980}}\) = 0.452 s.
So, time interval between the first and the second impacts = 1.3 – 0.452 = 0.848 s.
Time taken by the ball to reach the highest point after the first impact is,
t1 = \(\frac{1}{2}\) × 0.848 = 0.424 s
If the velocity of separation of the ball after the first impact is v, then
0 = v – gt1
or, v = gt1 = 980 × 0.424 = 415.5 cm ᐧ s-1
∴ Coefficient of restitution,
e = \(\frac{v}{V}\) = \(\frac{415.5}{442.7}\) = 0.94
Example 4.
A bullet of mass m hits a wooden block of mass M, suspended by a string of length l, and gets embedded in it. If the velocity of the bullet is v, find the angular displacement of the block.
Solution:
Let the velocity of the block-bullet system after impact be V and the angular displacement be θ [Fig.].
From the law of conservation of momentum,
mv = (M + m)V or,V = \(\frac{m v}{M+m}\)
Kinetic energy of the block-bullet system at position A
Example 5.
A boy of mass m1, standing on a smooth horizontal surface, throws a sphere of mass m2 parallel to the surface. After a time t, if the separation between them becomes x, then show that the work done by the boy in throwing the sphere = \(\frac{1}{2}\left(\frac{x}{t}\right)^2\left(\frac{m_1 m_2}{m_1+m_2}\right)\)
Solution:
Let v denote the velocity of the sphere, and V denote the velocity of the boy, due to reaction. From the law of conservation of momentum,
m1V = m2v …….. (1)
Separation between them after a time t
x = (V + v)t ……… (2)
From (1) and (2) we get,
\(\frac{m_2}{m_1} \nu\) + v = \(\frac{x}{t}\) or, v = \(\frac{x}{t} \cdot \frac{m_1}{m_1+m_2}\)
Similarly, V = \(\frac{x}{t} \cdot \frac{m_2}{m_1+m_2}\)
Work done in throwing the sphere
= sum of kinetic energies of the sphere and the boy
Example 6.
Two blocks m1 and m2 of masses 2kg and 5kg, respectively, are moving on a smooth plane along a straight line in the same direction, with velocities 10 m ᐧ s-1 and 3 m ᐧ s-1 respectively. The block m2 is situated ahead of the block m1. An ideal spring (k = 1120 N ᐧ m-1) is attached to the back of the block m2. Find the compression of the spring when m1 collides with m2.
Solution:
Let initial velocities of blocks m1 and m2 be v1 and v2 respectively. After collision, the two blocks combine and move with a velocity V [Fig.].
From the law of conservation of momentum,
m1v1 + m2v2 = (m1 + m2)V
or, 2 × 10 + 5 × 3 = (2 + 5)V
or, V = 5 m ᐧ s-1
Suppose the spring is compressed by x. From the law of conservation of energy we get
total energy before collision = total energy after collision
or, total kinetic energy of the blocks = kinetic energy of the combined blocks + potential energy of the compressed spring
i.e., \(\frac{1}{2} m_1 v_1^2\) + \(\frac{1}{2} m_2 v_2^2\) = \(\frac{1}{2}\)(m1 + m2)V2 + \(\frac{1}{2}\)kx2
or, 2 × (10)2 + 5 × (3)2 = 7 × (5)2 + 1120 × x2
∴ x2 = \(\frac{1}{16}\)
or, x = \(\frac{1}{4}\) = 0.25 m.
Example 7.
A ball moving at 9m ᐧ s-1, collides with an identical ball at rest. After collision, both the balls scatter at 300 with the initial direction of motion. Find the velocity of each ball after collision. Does the kinetic energy remain conserved in such a collision?
Solution:
Let mass of each ball = m. Velocities of the two balls before collision, u1 = 9 m ᐧ s-1 and u2 = 0 [Fig.].
Let their velocities after collision be v1 and v2 respectively.
From the law of conservation of the x and y components of momentum,
9m + 0 = mv1cos30° + mv2cos30° ……. (1)
and 0 = mv1sin30° – mv2sin30° …… (2)
From (1),
v1 + v2 = \(\frac{18}{\sqrt{3}}\) …… (3)
and from (2),
v1 – v2 = 0 ……. (4)
∴ Solving (3) and (4),
v1 = v2 = 3\(\sqrt{3}\) ᐧ s-1
Initial kinetic energy = \(\frac{1}{2} m u_1^2\) + \(\frac{1}{2} m u_2^2\) = \(\frac{1}{2}\)m(92 + 0) = \(\frac{81}{2}\)m
Final kinetic energy = \(\frac{1}{2} m v_1^2\) + \(\frac{1}{2} m v_2^2\) = \(\frac{1}{2}\)m(27 + 27) = 27m
Hence, kinetic energy does not remain conserved in this case.
Example 8.
Two identical blocks A and B, each of mass m, are connected to each other with a spring of length L [Fig.]. Force constant of the spring is k. The system is kept on a horizontal table. An identical third block C moving with a velocity y along the line AB collides with A elastically.
(i) What is the kinetic energy of the system A – B in the most compressed state?
(ii) What is the maximum compression of the spring?
Solution:
i) The velocity of block C, just before collision = v. As the collision of C with A is elastic, the block C comes to rest just after collision, and the block A gets the velocity v in the forward direction. Because of the spring between A and B, velocity of A gradually decreases, but that of B gradually increases, and the spring is compressed. As soon as velocities of the blocks A and B become equal, the spring attains maximum compression and the system A-B begins to move with a common velocity v1.
From the law of conservation of momentum,
mv = 2m × v1
or, v1 = \(\frac{\nu}{2}\)
∴ At maximum compression of the spring, kinetic energy of the system A-B
= \(\frac{1}{2}(2 m) v_1^2\) = \(\frac{1}{2}\) × 2m × \(\frac{v^2}{4}\) = \(\frac{1}{4} m v^2\)
ii) If the maximum compression of the spring is x, then potential energy of the compressed spring
= \(\frac{1}{2} k x^2\) = \(\frac{1}{2} m v^2\) – \(\frac{1}{4} m v^2\) = \(\frac{1}{4} m v^2\)
or, x2 = \(\frac{m}{2 k} v^2\)
or, x = v\(\sqrt{\frac{m}{2 k}}\)
Example 9.
A mass 2m is at rest and another mass m is moving with a velocity. An elastic collision takes place between them. Show that the mass m loses \(\frac{8}{9}\) part of its initial kinetic energy in this collision. [WBJEE ‘03]
Solution:
Let the velocity of mass m before collision = u and the velocity of mass m after collision = v1 . The velocity of mass 2m after collision = v2
From the law of conservation of momentum,
mu = mv1 + 2mv2 or, u = v1 + 2v2
or, u – v1 = 2v2
From the law of conservation of energy,
Initial kinetic energy of the mass m = \(\frac{1}{2} m u^2\)
Its kinetic energy after collision = \(\frac{1}{2} m v_1^2\) = \(\frac{1}{2} m \cdot \frac{1}{9} u^2\) = \(\frac{1}{18} m u^2\)
= \(\frac{1}{2} m u^2\) – \(\frac{1}{18} m u^2\) = \(\frac{8}{9} \times \frac{1}{2} m u^2\)
= \(\frac{8}{9}\) of its initial kinetic energy.
Example 10.
Two flat discs A and B are kept on a smooth horizontal table. The disc A moves with a velocity u and makes a perfectly elastic collision with B. If the mass of A is k times the mass of B, using conservation laws, find the fraction of kinetic energy of A transferred from A to B. Also prove that if the mass of B was k times the mass of A, the fraction of kinetic energy transferred would have been the same. [HS ‘02]
Solution:
Let mass of disc A be m1 and that of disc B = m2.
∴ m1 = km2
Let the velocities of A and B after collision be v1 and v2, respectively. From the law of conservation of momentum,
m1u + 0 = m1v1 + m2v2
or, km2u = km2v1 + m2v2 ∴ k(u – v1) = v2 …….. (1)
As the collision is perfectly elastic, the kinetic energy is also conserved.
Hence, in both cases, the same fraction of kinetic energy of A is transferred to B.
Example 11.
A small ball A travels in a vertical plane along a quarter of a circular path of radius 10 cm, as shown in
Fig., and hits another ball B of the same mass at rest.
Considering the collision to be elastic, and neglecting frictional force, find the velocity of each ball after the collision. [WBJEE ‘03]
Solution:
Let the mass of each ball be m and velocity of A just before it collides with B be u.
From the conservation of energy of A,
\(\frac{1}{2} m u^2\) = mgh
or, u = \(\sqrt{2 g h}\) = \(\sqrt{2 \times 980 \times 10}\) = 140 cm ᐧ s-1
In an elastic collision, two bodies of equal masses exchange their velocities. Hence, after collision, A will come to rest, and B will move with the velocity 140 cm ᐧ s-1.