Physics Momentum & Impulse

How to rearange: p=mv for v

v=p/m

How to rearange p=mv for mass

m=p/v

A 4.00kg model rocket is launched, shooting 50.0g of burned fuel from it exhaust at an average velocity of 625m/s. what is the velocity of the rocket after the fuel has burned? (ignore effects of gravity and air resistance.)

1.Make list

of known quantities: mass of rocket (m) = 4.00 kg, mass of fuel (mf) = 0.050 kg, velocity of fuel (vf) = 625 m/s. 2. Use conservation of momentum: initial momentum = final momentum. The final velocity of the rocket can be calculated using the conservation of momentum principle, resulting in a velocity of 7.91 m/s after the fuel has burned.

A thread holds two carts together on a frictionless surface. A compressed spring acts upon the carts holding them together. After the thread is burned, the 1.5kg cart moves with a velocity of 27 cm/s to the left what is the velocity of the 4.5kg cart?

First make a list. Known quantities: mass of cart 1 (m1) = 1.5 kg, velocity of cart 1 (v1) = -0.27 m/s, mass of cart 2 (m2) = 4.5 kg. Use conservation of momentum: initial momentum = final momentum. The final velocity of the 4.5 kg cart can be calculated, resulting in a velocity of 0.09 m/s to the right.

Two campers dock a canoe/ One camper steps onto the dock. This camper has a mass of 80.0kg and moves forward at 4.0m/s. With what speed and direction do the canoe and the other camper move if their combined mass is 110kg.

First make a list. Known quantities: mass of camper (m1) = 80.0 kg, velocity of camper (v1) = 4.0 m/s, combined mass of canoe and camper (m2) = 110 kg. Use conservation of momentum to find the velocity of the canoe and the other camper, resulting in a velocity of 2.91 m/s in the opposite direction.

A 0.105kg hockey puck moving at 48m/s is caught by a 75kg goalie at rest. With what speed does the goalie slide on the ice

1. Use Conservation of Momentum:

m_p v_p + m_g v_g = (m_p + m_g) v_f

Since v_g = 0, the equation simplifies to:

(0.105 \times 48) = (0.105 + 75) v_f

2. Solve for v_f:

5.04 = 75.105 v_f

v_f = \frac{5.04}{75.105} \approx 0.0671 \text{ m/s}

Final Answer:

The goalie slides at 0.067 m/s (or 6.71 cm/s).

A 35g bullet strikes a 5kg stationary wooden block and embeds itself in the block. The block and the bullet fly off together at 8.6m/s. What was the original velocity of the bullet.

1. Use Conservation of Momentum:

m_b v_b + m_w v_w = (m_b + m_w) v_f

Since v_w = 0, the equation simplifies to:

(0.035 \times v_b) = (0.035 + 5) \times 8.6

2. Solve for v_b:

0.035 v_b = 5.035 \times 8.6

0.035 v_b = 43.301

v_b = \frac{43.301}{0.035} \approx 1237.2 \text{ m/s}

Final Answer:

The bullet’s initial velocity was 1237 m/s.

What is the final velocity of the 1kg ball after a collision with a 0.5kg ball moving at 6 m/s and the 1kg ball moving at -12 m/s, given that the 0.5kg ball moves at -14 m/s afterwards?

1. Use Conservation of Momentum: m1v1_initial + m2v2_initial = m1v1_final + m2v2_final.

2. Known values: m1 = 0.5 kg, v1_initial = 6 m/s, v1_final = -14 m/s; m2 = 1 kg, v2_initial = -12 m/s.

3. Set up equation: (0.5 kg * 6 m/s) + (1 kg * -12 m/s) = (0.5 kg * -14 m/s) + (1 kg * v2_final).

4. Calculate: 3 - 12 = -7 + v2_final -9 = -7 + v2_final v2_final = -9 + 7 = -2 m/s. Final Answer: The velocity of the 1kg ball after the collision is -2 m/s.

What was the fullback's momentum before the collision with a 128kg tackle, given he weighs 95 kg and runs at 8.2 m/s?

Momentum (p) is calculated as mass (m) times velocity (v). For the fullback: p = m * v = 95 kg * 8.2 m/s = 779 kg·m/s.

What was the change in the fullback's momentum during the collision with a 128kg tackle, if he weighs 95 kg and runs at 8.2 m/s before both players end up with zero speed?

The change in momentum is calculated as the final momentum minus the initial momentum. Since both players end up at rest, the final momentum is 0. The initial momentum of the fullback is p_initial = 95 kg * 8.2 m/s = 779 kg·m/s. Thus, the change in momentum is: 0 - 779 kg·m/s = -779 kg·m/s.

What was the change in the tackle's momentum during the collision with a 95 kg fullback running at 8.2 m/s, if both players end up with zero speed?

The change in momentum is the final momentum minus the initial momentum. The initial momentum of the tackle is equal to the momentum of the fullback: p_initial = (95 kg * 8.2 m/s) + (128 kg * v_tackle_initial). Since both players end up at rest, the final momentum is 0. Therefore, the change in momentum for the tackle is equal to - (95 kg * 8.2 m/s), which is -779 kg·m/s.

What was the tackle's original momentum in a collision with a 95 kg fullback running at 8.2 m/s, if both players end up with zero speed?

The tackle's original momentum can be calculated using the principle of conservation of momentum: p_initial_fullback + p_initial_tackle = 0. For fullback: p_initial_fullback = 95 kg * 8.2 m/s = 779 kg·m/s. Therefore, p_initial_tackle = -779 kg·m/s (opposite direction).

Question: How fast was the 128 kg tackle moving originally in a collision with a 95 kg fullback running at 8.2 m/s, if both players end up with zero speed?

Answer: To find the tackle's original velocity (v_tackle_initial), use conservation of momentum: p_initial_fullback + p_initial_tackle = 0. The initial momentum of the fullback is p_fullback = 95 kg * 8.2 m/s = 779 kg·m/s. Therefore, for the tackle's momentum to balance, p_tackle = -p_fullback, giving v_tackle_initial = -p_fullback / 128 kg = -779 kg·m/s / 128 kg = -6.09 m/s.

What is the impulse that acted on a 25kg object moving at 12m/s after it collides and moves at 8m/s?

Impulse (J) is the change in momentum. First, calculate the initial momentum (p_initial = m * v_initial) = 25kg * 12m/s = 300 kg·m/s and final momentum (p_final = m * v_final) = 25kg * 8m/s = 200 kg·m/s. Impulse is then J = p_final - p_initial = 200 kg·m/s - 300 kg·m/s = -100 kg·m/s.

What was the initial speed of a 2575kg van running into a stationary 825kg compact car, if they move off together at 8.5m/s?

Using the conservation of momentum: ( m_{van} v_{van} + m_{car} v_{car} = (m_{van} + m_{car}) v_{final} ).\n( 2575 v_{van} + 825 * 0 = (2575 + 825) * 8.5 )\n( 2575 v_{van} = 3400 * 8.5 )\n( v_{van} = \frac{3400 * 8.5}{2575} \approx 13.53 , m/s ).