Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :



v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :


J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :



F = 0.8409 N
Hence, this is the required solution.
Answer:
1.503 J
Explanation:
Work done in stretching a spring = 1/2ke²
W = 1/2ke²........................... Equation 1
Where W = work done, k = spring constant, e = extension.
Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.
Substitute into equation 1
W = 1/2(26)(0.34²)
W = 13(0.1156)
W = 1.503 J.
Hence the work done to stretch it an additional 0.12 m = 1.503 J
We need to find the average speed of the ball during the motion of 1 m
In order to find that we took several reading and found following times to cover the distance of 1 m
t1 = 2.26 s
t2 = 2.38 s
t3 = 3.02 s
t4 = 2.26 s
t5 = 2.31 s
Now in order to find the average time we can write



So average time to cover the distance of 1 m by ball will be 2.45 s
here 3.02 s is not the average time but we can say it is the median of the readings of all possible values which we can not use in our calculation as average time
Answer:
the number of neutrons and protons in an atom
Explanation:
Given:
Final speed of mass A = Va
Final speed of mass B = Vb
Mass of A = Ma
Mass of B = Mb
Ma = 2 × Mb
By conservation of linear momentum,
0 = Ma × Va + Mb × Vb
0 = 2 × Mb × Va + Mb × Vb
Vb = - 2 × Va
Energy of the spring, U = 1/2 × k × x^2
1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
Ma × Va² + Mb × Vb² = 70
2 × Mb (-Vb/2)² + Mb × Vb² = 70
1/2 × Mb × Vb² + Mb × Vb² = 70
3/2 × Mb × Vb² = 70
Mb × Vb² = 140/3
= 46.7 J
Ma = 2 × Mb and Vb = - 2 × Va
Ma/2 × (4 × Va²) = 140/3
Ma × Va² = 70/3
Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J
Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J