A 4.0-kg particle is moving horizontally with a speed of 5.0 m/s when it strikes

Question

A 4.0-kg particle is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall. The particle rebounds with a speed of 3.0 m/s. What is the magnitude of the impulse delivered to the particle? 24 N middot s 32 N middot s 40 N middot s 30 N middot s 8.0 N middot s A 2.0 kg object moving with velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with a speed of 2.0 m/s in the same direction. How much kinetic energy is lost in this collision? 2.4 J 9.6 J 5.4 J 0.6 J 6.0J A 12-g bullet is fired into a 3.0-kg ballistic pendulum initially at rest and becomes was embedded in it. The pendulum subsequently rises a vertical distance of 12 cm. What speed of the bullet? 0.38 km/s 0.44 km/s 0.50 km/s 0.54 km/s 0.024 km/s A 10-g bullet moving 1000 m/s strikes and passes through a 2.0-kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To what maximum height will the block rise above its initial position? 78 cm 66 cm 56 cm 46 cm 37 cm

Explanation / Answer

12. (b) is the correct option.

Impulse can be given as:

Impulse (J)= change in momentum

Initial momentum:

pi= m×v = (4kg)×(5m/s) =- 20 kg-m/s { Towards vertical wall}

Final momentum:

pf= (4kg) × (3m/s) = 12 kg- m/s { away from wall}

Hence

J = 12- (-20) = 32 N.s

13. (c) is the correct option

Momentum is conserved

Initial momentum= final momentum

m1v1 +m2v2= (m1+m2)vf

(2)5 +(3)2 = (2+3)vf

Vf = 3.2 m/s

Initial kinetic energy :

(1/2) m1v12 + (1/2) m2v22

(1/2) (2)(5)2+(1/2)(3)(2)2 = 31 J

Final kinetic energy:

(1/2) (m1+m2)vf2 = (1/2)(5)(3.2)2= 25.6 J

Kinetic energy lost = 31 J- 25.6 J = 5.4 J

14. (a) is the correct option.

Applying conservation of energy :

Kinetic energy of system after impact = potential energy

let m= mass of bullet

M= mass of pendulum

1/2( m+M)v2 = U

1/2 (m+M)v2= (m+M)gh

i.e. v = 2gh

Now applying conservation of linear momentum

Let u be the initial velocity of bullet

mu = (m+M)v

mu = (m+M)2gh

By putting values we get u = 384.94 m/s or 0.38 km/s

15. (d) is the correct option

Momentum is conserved

m1v1 +m2v2= m1v1f +m2v2f

(10 g)(1000m/s) +0 = (10g)(400m/s) +(2000g)(v2f)

v2f= 3 m/s

Maximum height h = v2f2/2g

i.e. h = 46 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.