57. A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. It rebounds with a speed of 13.5 m/s. The contact time is 0.100 seconds. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.
(a) -16.7 N s; (b) -167 N
Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s
Find: (a) Impulse, (b) Force
(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)
Impulse = -16.7 kg•m/s = -16.7 N•s
where the "-" indicates that the impulse was opposite the original direction of motion.
(Note that a kg•m/s is equivalent to a N•s)
(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.
Impulse = F•t
F = Impulse/t = (-16.7 N s) / (0.100 s) = -167 N
where the "-" indicates that the impulse was opposite the original direction of motion. 58. A 4.0-kg object has a forward momentum of 20. kg•m/s. A 60. N•s impulse acts upon it in the direction of motion for 5.0 seconds. A resistive force of 6.0 N then impedes its motion for 8.0 seconds. Determine the final velocity of the object.
vf = 8.0 m/s
This question is best thought about conceptually using the principle that an objects momentum is changed when it encounters an impulse and the amount of change in momentum is equal to the impulse which it encounters.
Here an object starts with 20 units (kg•m/s) of momentum. It then encounters an impulse of 60 units (N•s) in the direction of motion. A 60-unit impulse will change the momentum by 60 units, either increasing or decreasing it. If the impulse is in the direction of an object's motion, then it will increase the momentum. So now the object has 80 units (kg•m/s) of momentum. The object then encounters a resistive force of 6.0 N for 8.0 s. This is equivalent to an impulse of 48 units (N•s). Since this impulse is "resistive" in nature, it will decrease the object's momentum by 48 units. The object now has 32 units of momentum. The question asks for the object's velocity after encountering these two impulses. Since momentum is the product of mass and velocity, the velocity can be easily determined.
p = m•v
vfinal = pfinal / m = (32 kg m/s) / (4.0 kg) = 8.0 m/s. 59. A 3.0-kg object is moving forward with a speed of 6.0 m/s. The object then encounters a force of 2.5 N for 8.0 seconds in the direction of its motion. The object then collides head-on with a wall and heads in the opposite direction with a speed of 5.0 m/s. Determine the impulse delivered by the wall to the object.
53 N•s
Like the previous problem, this problem is best solved by thinking through it conceptually using the impulse-momentum change principle.
Here the object begins with a momentum of 18 units (kg•m/s). The object encounters a force of 2.5 N for 8.0 seconds. This is equivalent to an impulse of 20 units (N•s). Since this impulse acts in the direction of motion, it changes the object's momentum from 18 units to 38 units. A final impulse is encountered when colliding with a wall. Upon rebounding, the object has a momentum of -15 units (kg•m/s). The -15 is the product of mass (3 kg) and velocity (-5 m/s). The "-" sign is used since the object is now moving in the opposite direction as the original motion. The collision with the wall changed the object's momentum from +38 units to -15 units. Thus, the collision must have resulted in a 53-unit impulse since it altered the object's momentum by 53 units. 60. A 46-gram tennis ball is launched from a 1.35-kg homemade cannon. If the cannon recoils with a speed of 2.1 m/s, determine the muzzle speed of the tennis ball.
62 m/s
Given: mball = 46 g = 0.046 kg; mcannon = 1.35 kg; vcannon = -2.1 m/s
Find: vball = ???
The ball is in the cannon and both objects are initially at rest. The total system momentum is initially 0. After the explosion, the total system momentum must also be 0. Thus, the cannon's backward momentum must be equal to the ball's forward momentum.
mcannon • vcannon = -mball • vball
(1.35 kg) • (-2.1 m/s) = (0.046 kg) • vball
vball = (1.35 kg) • (2.1 m/s) / (0.046 kg) = 61.63 m/s = ~62 m/s. 61. A 2.0-kg box is attached by a string to a 5.0-kg box. A compressed spring is placed between them. The two boxes are initially at rest on a friction-free track. The string is cut and the spring applies an impulse to both boxes, setting them in motion. The 2.0-kg box is propelled backwards and moves 1.2 meters to the end of the track in 0.50 seconds. Determine the time it takes the 5.0-kg box to move 0.90 meters to the opposite end of the track.
