A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for Δ t = 0.01 sec., the impulse of the impact force is : (take g = 10 m/s 2 )
Text Solution
Verified by ExpertsThe correct answer is:
B
v 1 = 
= 
= 
k 2 = 
k 1 ⇒ v 2 2 = 
v 1 2
∴ v 2 = 
= 
| Δ P| = |–mv 2 – (mv 1 )| = m |–v 2 – v 1 |
| Δ P| = 50 × 10 –3 × 
× 
= 
J = Δ P = 1.06 N-s.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The area of F-t curve is A, where 'F' is the force acting on one mass due to the other. If one of t…
A bullet of mass m moving vertically upwards instantaneously with a velocity 'u' hits the hanging b…
A ball of mass = 100 gm is released from a height h 1 = 2.5 m from the ground level and then reboun…
A system of two blocks A and B are connected by an inextensible massless strings as shown. The pull…
The area of F-t curve is A, where 'F' is the force on one mass due to the other. If one of the coll…
The given figure shows a plot of the time dependent force F x acting on a particle in motion along …