Question

A 2-kg pellet travels with velocity 60 m/s to the right when it collides with a 38-kg hanging mass which is initially at rest. After the collision, the pellet remains lodged in the hanging mass, i.e., it is a completely inelastic collision. The hanging mass (+pellet) then swings upward and reaches a maximum height hmax before swinging downward again. Assume that no external forces are present and therefore the momentum of the system is conserved.

1.What is the velocity vf of the hanging mass + pellet immediately after the collision? m/s Submit Answer Tries 0/3

2.What is the final kinetic energy Kf of the hanging mass + pellet immediately after the collision? J Submit Answer Tries 0/3

3.What is the maximum height hmax of the swinging hanging mass + pellet in centimeters?

Answer 1

1.
`v_(f)` = 5.45 m/s , 2. K = 326.73 J and 3. h = 152 cm

Step-by-step explanation:

R1. Let's use the conservation of the moment, for this we define a system formed by the two bodies, the pill plus the hanging mass,

Where the mass of the tablet (m = 2 kg) and the hanging mass (M = 38 Kg)

Initial, before crash

po = m v₀₁ + 0

Final, just after the crash

`p_(f)` = (m + M)
`v_(f)`

The moment is preserved

p₀ =
`p_(f)`

m v1o = (m + M)
`v_(f)`

`v_(f)` = m / (m + M) v1o

`v_(f)`= 2/(2+20) 60

`v_(f)` = 5.45 m/s

R2 The kinetic energy is given, in our case, after the collision

K = ½ (m + M)
`v_(f)`²

K = ½ (2 +20) 5.45²

K = 326.73 J

R3 Let's use the conservation of mechanical energy, after the crash. Let's look for energy at two points the lowest and the highest point

Lowest point

Em₀ = K = ½ (m + M)
`v_(f)`²

Highest point

`Em_(f)` = U = mg h

Em₀ =
`Em_(f)`

½ (m + M)
`v_(f)`² = (m + M) g h

h =
`v_(f)`² / 2g

h = 5.45²/2 9.8

h = 1.52 m (100cm / 1m)

h = 152 cm