A 6.0x10-2kg hollow racquetball with an initial speed of 18.6 m/s collides with a backboard. It rebounds with a speed of 4.6 m/s
1 answer:
Answer:
The total impulse that the wall imparts on the ball on the ball is 1.392 kg-m/s.
Explanation:
Given that,
Mass of the racquetball,
Initial speed of the ball, u = 18.6 m/s
Final speed of the ball, v = -4.6 m/s (the ball rebounds so it will be negative)
We need to find the total impulse that the wall imparts on the ball on the ball. We know that the change in momentum of an object. It is given by :
J = -1.392 kg-m/s
So, the total impulse that the wall imparts on the ball on the ball is 1.392 kg-m/s. Hence, this is the required solution.
You might be interested in
Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N