University:
Jackson College
Course:
PHY 231 | College Physics I
Academic year:

2024
Views:

150

Pages:

4
Author:

Leila Blankenship

MOMENTUM - Practice Exercises Formulas: p =m v F ~ net t=Δ p ^ or F ~ net =m a ~ v 1. Calculate the momentum of a 4.0 kg object travelling at a velocity of 12.0 m/s east. \vec{P} = m\vec{v} = (4)(12)[E] = \underline{48 kg\cdot m/s [E]} 2. A 5.0 kg object has a momentum of 25.0 kg·m/s west. What is its velocity? \vec{v} = \frac{\vec{p}}{m} = \frac{25[W]}{5} = \boxed{5.0 m/s [W]} ✓ 3. An object has a velocity of 8.0 m/s south and a momentum of 36.0 kg·m/s south. What is the mass of the object? m = \frac{\vec{p}}{\vec{v}} = \frac{36[S]}{8[S]} = \underline{4.5 kg} ✓ 4. An object has a velocity of 2.0 m/s east and momentum of 29 kg·m/s east. What is the magnitude of the weight of the object? M = \frac{\vec{p}}{\vec{v}} = \frac{29[E]}{2[E]} = 14.5 kg ← MASS Fg = (14.5)(9.81) = \underline{1.4 \times 10^2 N} 5. A 6.6 N object is travelling at a velocity of 3.0 m/s north. What is the momentum of the object? (Express answer in kg·m/s.) m = \frac{F}{a} = \frac{6.6}{9.81} = 0.67278 kg \vec{p} = m\vec{v} = (0.67278)(3 m/s [N]) = \underline{2.0 kg·m/s [N]} ✓ CASTLE ROCK RESEARCH Lesson 1 - Practice Exercises A 7.0 kg object travels 2.6 m west in 1.1 s. If velocity is uniform, what is the momentum of the object? v = d/t = 2.6/1.1 = 2.36 m/s [W] P = mv = (7 kg)(2.36 m/s [W]) = 16.54 ≈ 17 kg·m/s [W] ✓ A 5.0 kg object is dropped from a height of 2.5 m above the floor. What is the object's momentum after 0.25 s? P = mv v_i = 0 v_f = v_i + at = (9.81)(0.25) = 2.4525 m/s P = (5 kg)(2.4525 m/s) = 12.26 ≈ 12 kg·m/s [down] An average net force of 17.0 N acts east on an object for 2.5 x 10^-2 s. What is the impulse? impulse = Ft = (17 N)(2.5 x 10^-2 s) = 0.425 ≈ 0.43 N·s [E] ✓ An average net force of 11.2 N acts west on an object producing an impulse of 7.00 N·s west. How long did the force act on the object? t = impulse/F = 7.00 N·s [W] / 11.2 N [W] = 0.625 s A 26.3 kg object is traveling at 21.0 m/s north. What average net force is required to bring this object to a stop in 2.60 s? impulse = ΔP = P_f - P_i Ft = -mv F_net = -mv/t = -(26.3 kg)(21.0 m/s [N]) / 2.60 s = -212 N Not for Reproduction Physics 12 SNAP MOMENTUM - Practice Exercises An average net force of 31.6 N south is used to accelerate a 15.0 kg object uniformly from rest to 10.0 m/s. How long was the acceleration? F_net * t = ΔP = P_f - P_i t = mv/F = (15)(10)/31.6 = 4.75 s ✓ An average net force of 25.0 N acts north on an object for 7.20 x 10^-1 s. What is the change in momentum of the object? ΔP = impulse = FΔt = (25 N)(7.2 x 10^-1 s) = 18.0 kg·m/s [N] ✓ A 5.00 kg object accelerates uniformly from rest to a velocity of 12.0 m/s east. What is the impulse on the object? ΔP = P_f - P_i = mv = (5)(12) = 75 N·s [E] ✓ An average net force caused an 11.0 kg object to accelerate uniformly from rest. If this object travels 26.3 m west in 3.20 s, what is the change in momentum of the object? ΔP = P_f - P_i = mv v_f = 2d/t = 2(26.3)/3.2 = 16.4375 m/s [W] ΔP = (11)(16.4375) = 181 kg·m/s [West] ✓ A 3.0 kg object is dropped from a height of 6.5 m. How far has the object fallen when its momentum is 6.0 kg·m/s down? ΔP = P_f - P_i = mv v_f = 6/3 = 2 m/s [down] v_f² = v_i² + 2ad d = v_f²/2a = (2)²/(2)(-9.81) = 0.203 m [down] CASTLE ROCK RESEARCH