Momentum & Rotational Motion

As shown in the diagrams below, a lump of clay travels horizontally to the right toward a block at rest on a frictionless surface. Upon collision, the clay and the block stick together and move to the right. Compared to the total momentum of the clay and the block before the collision, the momentum of the clay-block system after the collision is:

• can't be determined

• greater

• less

• the same

Tags

NGSS.HS-PS2-2

If a 50-kg box is sliding on a frictionless surface with a certain velocity, and 10 pigeons land on it, what happens to the velocity of the box?

• increases

• decreases

• stays the same

• impossible to tell

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-2

A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble for the same amount of time. How does the rate of change of the boulder's momentum compare to the rate of change of the pebble's momentum?

• greater than

• less than

• equal to

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-2

A propeller with radius r rotates with constant angular velocity. What happens to its tangential velocity & tangential acceleration if the radius & velocity are doubled?

• Both tangential velocity and tangential acceleration doubled.

• The tangential velocity is doubled while tangential acceleration remains the same.

• Both tangential velocity and tangential acceleration are quadrupled.

• The tangential velocity is quadrupled while the tangential acceleration remains the same.

Children are playing in a roundabout and they want to spin faster. Assume no external forces are applied, what could be done to make it rotate faster?

• Add more children at the edge of the roundabout.

• Have all of them sit instead of standing up.

• Have children with larger weights move to the center of the roundabout.

• Add more children to the center of the roundabout.

Tags

NGSS.HS-PS2-1

A 5500-kg freight truck accelerates from 4.2 m/s to 7.8 m/s in 15 seconds by the application of a constant force. A) What impulse is experienced? B) How large of a force is exerted?

Answer explanation

A)

J = Δp

J = (5500)(7.8) - (5500)(4.2)

J = 19,800 kg*m/s

B)

J = Ft

19,800 = F(15)

F = 1320 N

Tags

NGSS.HS-PS2-1

A 0.045-kg golf ball is dropped from a height of 1.8 m and comes back up at a height of 1.2 m. A) What is the velocity of the ball right before it hits the ground? B) What is the rebound velocity? C) What impulse did the ball experience?

Answer explanation

A)
Potential Energy = Kinetic Energy

U = KE or PE = KE

(0.045)(9.8)(-1.8) = 1/2(0.045)v^2

v = -5.94 m/s

B)

KE = U or KE = PE

1/2 (0.45)v^2 = (0.045)(9.8)(1.2)

v = 4.85 m/s

C)

J = Δp

J = (0.045)(4.85)-(0.045)(-5.94)

J = 0.49 kg*m/s

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-2

NGSS.HS-PS3-1

A 1-kg firecracker initially at rest splits into two pieces. Piece #1 has a mass of 0.25 kg. Piece #2 moves with a velocity of 12 m/s to the right. A) What is the mass of the second piece? B) What is the velocity of the first piece?

Answer explanation

A)

Total mass = m₁ + m₂

1 = (0.25) + m₂

m₂ = 0.75 kg

B)

P_o = P_f

0 = (0.25)V₁ + (0.75)(12)

V₁ = -36m/s

Tags

NGSS.HS-PS2-2

A rugby player tackles an object. The collision is graphed in the diagram below. (A) What is the total impulse experienced by the rugby player? (B) If the 40-kg player is running forward at 7.0 m/s when she makes contact with the dummy, what is the player’s velocity at the end of the 7s impact?

Answer explanation

A)

1: 1/2(1)(6) = 3

2: (2)(6) = 12

3: 1/2(2)(6) = 6

4: 1/2(2)(-2) = -2

Total impulse: 19 kg*m/s

B)

J = Δp

19 = (40)V_f - (40)(7)

V_f = 7.48 m/s

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-2

A ceiling fan with a 0.15m radius starts from rest and completes 5 revolutions in 5s. A table fan with half the radius of the ceiling fan has half the angular acceleration as the ceiling fan. What is the tangential acceleration of the table fan?

Answer explanation

To find the tangential velocity of the table fan, we need to get the value of the angular velocity and angular acceleration of the ceiling fan using the following equations:

ω = ΔΘ/Δt

α = Δω/Δt

ΔΘ = 5 revolution = 5(2π) - 10π

ΔΘ =ω_ot + 1/2αt²

10π = 0 + 1/2α(5)²

α_tf = 2.51 rad/s² (angular acceleration of the table fan)

The tangential acceleration of the table fan can be solved using the following equation:

a_T = (r/2)(α_tf)
Remember, the table fan has half the radius of the ceiling fan

a_T = (0.15/2)(2.51) = 0.19 m/s²

Tags

NGSS.MS-PS2-1

NGSS.MS-PS2-2