A block of mass m_1 = 19 kg slides along a horizontal surface (with friction, mu

Question

A block of mass m_1 = 19 kg slides along a horizontal surface (with friction, mu_k = 03T) a distance d = 2.8 m before striking a second block of mass m_2 = 725 kg. The first block has an initial velocity of v = 7.75 m/s. m_1 =19 kg m_2 = 725 kg = 037 d = 2.8 m v= 7.75 m/s Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s? A numeric value is expected and not an expression. v_2 How far does block two travel, d_2 in meters, before coming to rest after the collision? A numeric value is expected and not an expression. d_2

Explanation / Answer

given
m1=19 kg
uk=0.37
d=2.8m
m2=7.25kg
v=7.5m/s

acceleration a = µk*g = 0.37 * -9.8m/s² = -3.626 m/s²

v² = u² + 2as = (7.5m/s)² - 2 * 3.626m/s² * 2.8m = 35.95 m²/s²
v = 5.99 m/s --------- impact velocity

Now conserve momentum:
19kg * 5.99m/s + 7.25kg * 0m/s = 19kg * 0m/s + 7.25kg * V
V = 15.69 m/s --------------------------(a)

(b) 0 = V² + 2ad = (15.69m/s)² - 2 * 3.626m/s² * d2
246.176=7.252*d2
d2=33.94 m

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