Answer to Question #268268 in Physics for Lili

Question

In the figure, block 1 of mass m₁ slides from rest along a frictionless ramp from height h = 2.3 m and then collides with stationary block 2, which has mass m₂ = 4m₁. After the collision, block 2 slides into a region where the coefficient of kinetic friction μ_k is 0.6 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?

The tolerance is in the 2nd significant digit.

Answer 1

(a) $mgh=mv^2/2\to v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot2.3}=6.71\ (m/s)$

$mv=mv_1+4mv_2\to6.71=v_1+4v_2$

$mv^2/2=mv_1^2/2+4mv_2^2/2\to 45=v_1^2+4v_2^2$

we have $v_2=2.68\ (m/s)$

$F=4ma\to \mu 4mg= 4ma\to a=0.6\cdot9.8=5.88\ (m/s^2)$

$s=\frac{v^2-v_0^2}{2a}=\frac{0^2-2.68^2}{2\cdot (-5.88)}=0.61\ (m)$ . Answer

(b) $mgh=mv^2/2\to v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot2.3}=6.71\ (m/s)$

$mv=(m+m_1)v_1\to v_1=mv/(m+4m)=6.71/5=1.34\ (m/s)$

$F=5ma\to\mu 5mg=5ma\to a=\mu g=0.6\cdot 9.8=5.88\ (m/s^2)$

$s=\frac{v^2-v_0^2}{2a}=\frac{0^2-1.34^2}{2\cdot (-5.88)}=0.15\ (m)$ . Answer