Question

Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a speed of 3.5 m/s. After the collision, the second ball is at rest.
A) What is the final speed of the first ball?
B) What is the kinetic energy of the first ball before the collision?
C) What is the kinetic energy if the second ball after the collision?

Answer 1

Answer:

(a) 3.5 m/s

(b) 0 J

(c) 0 J

Explanation:

mass of each ball, m = 0.4 kg

initial velocity of first ball, u1 = 0

initial velocity of second ball, u2 = 3.5 m/s

final velocity of second ball, v2 = 0

(a) let the final velocity of first ball is v1.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.4 x 0 + 0.4 x 3.5 = 0.4 x v1 + 0.4 x 0

v1 = 3.5 m/s

Thus, the final velocity of the first ball after collision is 3.5 m/s

(b) kinetic energy of first ball before collision

K1 = 0.5 x m1 x u1^2 = 0

(c) Kinetic energy of second ball after collision

K2 = 0.5 x m2 x v2^2 = 0

Answer 2

Explanation:

Mass of two soccer balls, $m_1=m_2=0.4\ kg$

Initial speed of first ball, $u_1=0$

Initial speed of second ball, $u_2=3.5\ m/s$

After the collision,

Final speed of the second ball, $v_2=0$

(a) The momentum remains conserved. Using the conservation of momentum to find it as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_1$ is the final speed of the first ball

$0.4\times 0+0.4\times 3.5=0.4v_1+0.4\times 0$

$0.4\times 3.5=0.4v_1$

$v_1=3.5\ m/s$

(b) Let $E_1$ is the kinetic energy of the first ball before the collision. It is given by :

$E_1=\dfrac{1}{2}mu_1^2$

$E_1=\dfrac{1}{2}\times 0.4\times 0$

It is at rest, so, the kinetic energy of the first ball before the collision is 0.

(c) After the collision, the second ball comes to rest. So, the kinetic energy of the second ball after the collision is 0.

Hence, this is the required solution.