Class Management Help Chapter 9- Linear Momentum Begin Date: 10/26 017 10:00:00

Question

Class Management Help Chapter 9- Linear Momentum Begin Date: 10/26 017 10:00:00 AM--Due Date: 11/7/2017 11:59:00 PM End Date: 11/7/2017 11:59:00 PM (10%) Problem 7: A block of mass m,-12 kg slides along a horizontal surface (with friction. ,-0.22) a distance d = 2.65 m before striking a second block of mass m2 = 6.75 kg. The first block has an initial velocity of v 9.25 m/s. Randomized Variables mi 12 kg m2-6.75 kg 0.22 d= 2.65 m v=9.25 m/s Otheexpertta.com 50% Part(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s? Grade Summary Potential Submissions V1- 0% 100% coso cotan asin0 sinO tan() | ( acoso 4 5 6 atan acotan sinhO cosh tah cotanho Attempts remaining: 6 (1%) per attempt) detailed view Degrees O Radians 0 Submit Hint I give up Hints: 0% deduction per hit. Hints remaining: Feedback: 1% deduction per feedback. 50% Part (b) How far does block two travel. ds in meters, before coming to rest after the collision? All enment 2017 Fenert TA. TJ.C

Explanation / Answer

here,

a)

accelration due to friction , a = - u * g = - 0.22 * 9.81 = - 2.16 m/s^2

let the speed of block 1 before the collison be u1

using the third equation of motion

v^2 - u^2 = 2 * a * d

u1^2 = 9.25^2 - 2 * 2.16 * 2.65

u1 = 8.61 m/s

let the speed of 2 after the collison be v2

using the conservation of momentum

m1 * u1 + 0 = 0 + m2 * v2

12 * 8.61 = 6.75 * v2

solving for v2

v2 = 15.3 m/s

b)

let the distance travelled before coming to rest after the collison be s

using third equation of motion

v^2 - u^2 = 2 * a * s

15.3^2 = 2 * 2.16 * s

s = 54.2 m

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---