Question

A ball of mass 0.5 kg moving with a velocity 5 ms¹ hits a wall normally and rebounds with half of the initial velocity. If the ball is in contact with the wall for 0.5 s, the force exerted by the wall on it is

(a) 3.75 m
(c) 7.5 N
(b) 3.75 x 10-2 N
(d) 0.375 N​

Answer 1

Given :

• Mass (m) = 0.5 Kg
• Initial velocity (u) = 5m/s²
• Final velocity (v) = -2.5m/s²

[Since the body rebounds]

• Contact time (t) = 0.5s

To find :

• Force exerted (F)

Solution :

We know :-

› Chane in momentum (∆P) = mv - mu

Substituting :-

› ∆P = 0.5(-2.5) - 0.5(5)

› ∆P = -1.25 - 2.5

∴ ∆P = -3.75 Kgm/s

We know :-

› Force = Change in momentum/Time taken

Substituting :-

› Force = -3.75/0.5

∴ Force = -7.5 N

Or, F = Ma

Where :-

• F = Force
• M = Mass
• A = Acceleration

[v - u/t]

Substituting :-

› F = 0.5 × (-2.5 - 5)/0.5

› F = 0.5 × (-7.5)/0.5

› F = 0.5 × -15

∴ F = -7.5N

∴ Force exerted on the wall is 7.5 N in the opposite direction of motion.

Answer 2

Answer:

Force exerted by wall on the ball during contact = 7.5N

Explanation:

Given,

Mass of the ball = m = 0.5 kg

Velocity with which it hits the wall = u = 5 m/s

Velocity after rebounding = v = 1/2 of initial velocity = 1/2 * 5 = 2.5 m/s

vector u = 5i

vector v = -2.5i ( -sign due to opposite direction)

Now, Change in momentum of ball when it is in contact with the wall

vector p = m(vector v - vector u)

vector p = 0.5*[-2.5i - 5i]

vector p = 0.5(-7.5i) [ -sign indicates motion of ball away from wall]

p = 0.5 * 7.5

p = 3.75

Force exerted by wall on the ball during contact = F = p/t

Duration of contact = 0.5 s

Hence, F = 3.75/0.5 = 7.5 N

Force exerted by wall on the ball during contact = 7.5N